Physics midterm question: cargo-loaded truck vs. truck-loaded cargo

Physics 205A Midterm 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

A Physics 205A student draws a (correct) free body diagram for a 600 kg cargo load resting on a stationary 11,000 kg truck with these two forces[*]:

Weight force of Earth on cargo load (5,800 N, downwards),

Normal force of truck on cargo load (5,800 N, upwards).

This student additionally claims that "this [free body diagram for the cargo load] would change if the truck was on top of the cargo load." Discuss why both the magnitude and direction of the normal force of truck on the cargo load would change if the truck were instead resting on top of the cargo load, and how you know this. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.


[*] waiferx.blogspot.com/2017/10/physics-midterm-question-proposed-test.html.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram(s), and discusses/demonstrates that when the truck is on top of the cargo load:
  1. the truck has two vertical forces acting on it:

     Weight force w of Earth on truck (m_truck·g = 107,800 N, downwards),
     Normal force N of cargo load on truck (107,800 N, upwards),

     and since there is no vertical motion, these two vertical forces must be equal in magnitude due to Newton's first law; and
  
  
  2. from Newton's third law, these two forces must have equal magnitudes and opposite directions:

     Normal force Nof cargo load on truck (107,800 N, upwards),
     Normal force Nof truck on cargo load (107,800 N, downwards),

     such that the normal force of truck on cargo load for the case where the truck is on top of the cargo load is both different in magnitude (107,800 N vs. 5,800 N) and direction (downwards vs. upwards) compared to the normal force of truck on cargo load in the case where the cargo load was on top of the truck.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Typically application of Newton's third law is problematic, or only implied, but still discusses how the normal force of truck on the cargo load is both different in magnitude and direction than in the previous case.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Some substantive attempt at analyzing forces using Newton's laws and free-body diagrams.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at using Newton's laws and free-body diagrams.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic attempt at using Newton's laws and free-body diagrams.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 17 students
r: 16 students
t: 18 students
v: 4 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 1995):

Physics midterm question: pulled box pulling on table underneath

Physics 205A Midterm 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

A Physics 205A student applies a force (magnitude of 32 N, directed to the left) to pull on a rope attached to a 12.0 kg box, which moves at constant speed to the left across a fixed, stationary table. Discuss why both the magnitude and direction of the kinetic friction force of the box on the table would also be 32 N, directed to the left. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram(s), and discusses/demonstrates:
  1. the box has two horizontal forces acting on it:

     Tension force T of student on box (32 N, to the left),
     Kinetic friction force  f_k of table on box (32 N, to the right),

     and since the box has a constant velocity ("constant speed to the left"), these two horizontal forces must be equal in magnitude due to Newton's first law; and
  
  
  2. from Newton's third law, these two forces must have equal magnitudes and opposite directions:

     Kinetic friction force  f_k of table on box (32 N, to the right),
     Kinetic friction force  f_k of box on table (32 N, to the left).

• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Some substantive attempt at analyzing forces using Newton's laws and free-body diagrams. Typically discusses Newton's first law for the forces acting on the box, but subsequent discussion of Newton's third law is omitted or merely implied.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at using Newton's laws and free-body diagrams.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic attempt at using Newton's laws and free-body diagrams.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 15 students
r: 5 students
t: 14 students
v: 15 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 6900):
Another sample "p" response (from student 2533; note that the static friction force of the ground on table pointing to the right is denoted as a tension force):
A sample "t" response (from student 1995), discussing Newton's first law for the box, demonstrating that the kinetic friction force of the table on the box points to the right with a magnitude of 32 N:

Physics midterm question: faster vertically swung ball

Physics 205A Midterm 1, fall semester 2019
Cuesta College, San Luis Obispo, CA

A Physics 205A student twirls a ball attached to a string in a vertical circle with constant speed. When the ball is swinging through the lowest part of the circle, the tension force on the ball is 1.8 N. Discuss why the magnitude of the tension force on this ball would have a larger magnitude if the ball had a faster constant speed while swinging through the lowest part of the circle (with the same radius). Explain your reasoning by using free-body diagram(s), the properties of forces and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram, and discusses/demonstrates:
  1. while the ball is swinging through the lowest part of the circle, there are two forces acting on the ball:

     Tension force T of hand on the ball (originally 1.8 N, upwards),
     Weight force w of Earth on the ball (constant magnitude of m⋅g, downwards);

  2. Newton's second law for uniform circular motion applies, such that while the ball is swinging through the lowest part of the circle, the net force ΣF (magnitude m⋅v²/r) must point in towards the center of the circular motion--which is vertically upwards; and
  3. increasing the speed v (while m and r are constant) would increase the required upwards net force for uniform circular motion, such that the upwards tension force will be greater than the original 1.8 N
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Some substantive attempt at applying Newton's second law for uniform circular motion.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying Newton's first law or third law for uniform circular motion.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic application of Newton's laws.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01duCk
p: 21 students
r: 11 students
t: 9 students
v: 9 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 1478):

Physics quiz question: comparing directions and magnitudes of forces

Physics 205A Quiz 3, fall semester 2019
Cuesta College, San Luis Obispo, CA

"A 2019 Chevrolet Silverado LT Trail Boss, New York International Auto Show"
Kevauto
commons.wikimedia.org

A cargo load (of unknown mass) rests on the bed of a 10,660 kg stationary truck[*]. Newton's __________ law tells you that these two forces are equal in magnitude and opposite in direction:

Normal force of cargo load on the truck.
Normal force of ground on the truck.

(A) first.
(B) second.
(C) third.
(D) (These forces are not equal in magnitude and/or opposite in direction.)

[*] chevrolet.com/commercial/silverado-chassis-cab.

Correct answer (highlight to unhide): (D)

The truck has three forces acting on it:

Weight force of Earth on the truck (m_truck·g = 1.04×10⁵ N, downwards).
Normal force of cargo load on the truck (downwards).
Normal force of ground on the truck (upwards).

Because the truck is stationary, the magnitudes of the two downward forces added together must equal the magnitude of the upwards force, due to Newton's first law. So the normal force of cargo load on the truck and the normal force of ground on the truck, while being opposite in direction, cannot be equal in magnitude (the difference in magnitudes being equal to the weight of the truck).

Sections 70854, 70855
Exam code: quiz03Ch3V
(A) : 22 students
(B) : 1 student
(C) : 9 students
(D) : 22 students

Success level: 41%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.65

Physics quiz archive: vectors, projectile motion, forces

Physics 205A Quiz 3, fall semester 2019
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855
Exam code: quiz03Ch3V



Sections 70854, 70855 results

0- 6 :	* [low = 6]
7-12 :	***
13-18 :	**************
19-24 :	******************* [mean = 21.7 +/- 5.8]
25-30 :	***************** [high = 30]

Online reading assignment: applications of Newton's laws (friction)

Physics 205A, fall semester 2019
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters on applications of Newton's laws (emphasizing static and kinetic friction).


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"There can only be two types of motion: constant and changing. There also only two types of net forces: zero and non-zero."

"Newton's third law of motion isn't necessarily about motion or net force, but the properties of force itself."

"We can use the mnemonic device 'POF-OST-ITO' in order to test if Newton's third law applies or not. POF stands for 'pair of opposite forces,' OST is 'of same type,' ITO 'involving two objects.'"

"Newton's third law relates only two forces of the same type acting on two different objects. However Newton's first law deals with two forces acting upon one object. Additionally, if the POF-OST-ITO test fails, then it is not a N3 scenario, but it is not yet safe to assume it is a N1 scenario either, until you do further investigation."

"Newton's second law shows how the acceleration depends on both the net force and the mass. The magnitude of acceleration is proportional to the net force acting on the object, and inversely proportional to the mass. The net force includes only the forces that the environment exerts on the object of interest. The kinetic friction force opposes the relative sliding motion."

"Everything is effected on by gravity. If something is touching a surface, it will have a normal force acting on it."

"The normal force can only exist when two surfaces are making contact with one another. If two objects are not making contact, then there is no normal force."

"Friction is another component when dealing with the motions of an object. It is a force that is parallel to the surface an object is moving on. Static friction is what makes an object remain stationary even when a force is applied, and only moves when the applied force is slightly greater than the maximum force of static friction."

"Static friction force comes into play when pulling on an object at rest, and if the force applied is small enough static friction will cancel out the applied force resulting in no movement."

"That an object initially at rest interacts with a surface through static friction. As the object moves, after overcoming the maximum static friction, the resistance turns into kinetic friction. Friction is a parallel and normal force is perpendicular to the surfaces in contact."

"Static friction is the friction that a surface has on a object that is resting on the surface and is directly proportional to the normal force on said object. Next, there is kinetic friction which is friction that occurs between two sliding surfaces. Also kinetic friction is typically less than the static friction."

"Static and kinetic frictional forces seem pretty straightforward. Static refers to when there is friction acting on an object but the object is not moving. Kinetic is when there is friction acting on an object but the object is moving."

"Was a little lost, so there isn't much I understand."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"I feel like I understand Newton's laws from the textbook. However, when it comes to applying the concepts to a free body diagram, I start getting confused with the direction of the forces and how they cancel out especially when things are stacked on top of each other."

"I still have trouble distinguishing between Newton's first law and second law."

"I was slightly confused as to why the pulleys and ropes would have equal tensions for the different worksheet examples in class. But after further research I understood that neither the mass nor acceleration of gravity changed, therefore the tension forces stayed the same."

"I get confused on how to determine which law is being used between Newton's first law or third law. I also don't understand the difference between static and kinetic friction forces and the magnitudes associated with them."

"The difference between static friction and kinetic friction aren't clear to me."

"The coefficient of friction and what it stands for might need some clarification for me."

"I would like some more work in class with problems on kinetic friction and static friction. I understand that static friction is friction for stationary objects, like a box on the ground, and that kinetic friction is friction for moving objects, like a box sliding across the ground. However, I'm still unsure of how this would work in real-world situations."

"I don't know how to put friction in an equation or how to use it to solve for something."

"Why are there no SI units for static/kinetic friction coefficient?"

"For the most part it is pretty easy to visualize friction, since we have all experienced it before."

"Didn't really find anything confusing."

"I understood most of it conceptually, I think it could get more confusing when these concepts are applied in a problem."

"Most of it."

What is the meaning of the "normal" in the "normal force?"

"The perpendicular force applied between two objects contacting each other."

"It just means the force is perpendicular to the surface."

"The 'normal' force refers to the perpendicular direction with respect to the surface."

"Normal means 'perpendicular.'"

"There will almost always be normal force on any given objects in contact, regardless of its state or location? That's what makes it 'normal,' right?

The SI (Système International) units of the static friction coefficient µ_s and the kinetic friction coefficient µ_k are:

" Unitless."

The coefficients because the units cancel."

"I do not believe these coefficients have units."

"Newtons?"

"kg·m²·s²?"

"I'm not entirely sure."

Identify the magnitude of the static friction force f_s for each of the following situations of a box that is initially stationary on a horizontal floor. (Only correct responses shown.)

No external horizontal forces applied to it, so it remains stuck to the floor:
f_s = 0. [77%]

An external horizontal force applied to it, but still remains stuck to the floor:
f_s = some value between 0 and µ_s·N. [83%]

An external horizontal force applied to it, at the threshold of nearly becoming unstuck:
f_s = µ_s·N. [72%]

Identify the magnitude of the kinetic friction force f_k for each of the following situations of a box that is already sliding across a horizontal floor. (Only correct responses shown.)

No external horizontal forces applied on it, so it slows down:
f_k = µ_k·N. [21%]

An external horizontal force applied in the forward direction, but not enough to keep the box going so it still gradually slows down:
f_k = µ_k·N. [23%]

An external horizontal force applied in the forward direction, just enough to keep the box going at a constant speed:
f_k = µ_k·N. [43%]

An external horizontal force applied in the forward direction, enough to gradually increase the speed of the box:
f_k = µ_k·N. [58%]

An external horizontal force applied in the backwards direction, such that the box slows down:
f_k = µ_k·N. [32%]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"Can we go over the friction coefficients? Please explain static friction force and kinetic friction force in detail in class! I could use some review on the equations for the magnitudes of static and kinetic frictional force. Is the friction coefficient different between objects that do not appear smooth. For example, is the constant for ice on ice different than say, bumpy ice on ice?" (Yes, the coefficient will be larger for rougher surfaces than for smoother sliding surfaces.)

"Are there ways to increase of decrease friction?" (Change the smoothness/roughness of the surfaces, or add a lubricant (which is "smooth" on a molecular level; long hydrocarbon chains can align with each other to roll like logs) between the two surfaces.)

"What is it that causes friction? Is it simply objects hooking onto each other at microscopic levels?"

"Does static friction force become kinetic friction force once the object starts moving?" (Yes, once the object is already unstuck.)

"What I found confusing was that static friction force and the applied force on a object are directly proportional, except for the fact that static friction has a maximum amount, as opposed to applied force which can keep increasing indefinitely. If the two are directly proportional, shouldn't the static friction force increase as long as the applied force increases, or shouldn't the applied force also have a maximum if the static friction has one as well?" (You can arbitrarily exert any amount of applied force on an object, but the static friction force will have a maximum amount because at some point your applied force will "unstick" the object so it will then begin to move. I suppose if you stuck together the two surfaces magic superglue, then you could exert an infinite amount of applied force, and the static friction force would then also be infinite, provided the magic superglue still holds the object stationary.)

"I hope I'm understanding this correctly! It seems to make sense so that's either a really good sign or a really bad sign."

"Definitely need some review, this weekend fried my brain."

"I don't like this :("

Physics quiz question: energy changes of hill-sliding student

Physics 205A Quiz 5, fall semester 2018
Cuesta College, San Luis Obispo, CA

A Physics 205A student on a cardboard sheet slides down a grassy slope starting from rest, and has a final speed of 0.80 m/s. Consider the Physics 205A student and cardboard sheet as a single 75 kg object. Friction is not negligible. Ignore drag.

For this process, the decrease in the Physics 205A student and cardboard sheet's gravitational potential energy is __________ the increase in translational kinetic energy.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (C)

The energy transfer-balance equation is given by:

W_nc = ∆KE_tr + ∆PE_grav + ∆PE_elas,

where ∆PE_elas = 0, as there is no spring involved in this process.

Just looking at the two remaining terms on the right-hand side of the energy transfer-balance equation, for the change in translational kinetic energy:

∆KE_tr = (1/2)·m·(v_f² – v₀²),

and since the speed is increasing, v_f is faster than v₀, and so KE_tr is increasing (∆KE_tr is positive).

Also for the change in gravitational potential energy:

∆PE_grav = m·g·(y_f – y₀),

and since y_f is lower than y₀, then PE_grav decreases (∆PE_grav is negative).

On the left-hand side of the energy transfer-balance equation, the work done by kinetic friction against the student and cardboard sheet is negative, as the kinetic friction force points up along the slope, while the displacement points down along the slope, such that:

W_nc = ∆KE_tr + ∆PE_grav,

with the ± signs as noted for each term:

(–) = (+) + (–),

and so the decrease in gravitational potential energy (negative change) must be greater than the increase in translational kinetic energy (positive change) on the right-hand side of the energy transfer-balance equation to be equal to the negative non-conservative work done by kinetic friction.

(Note that the normal force does no work on the student and cardboard sheet, as this force is perpendicular to the displacement, which points down the slope. Also we do not need to calculate the work done by the weight force on the student and cardboard sheet in this energy transfer-balance equation, as this is a conservative force that is already included in the gravitational potential energy term on the right-hand side of the equation.)

Sections 70854, 70855
Exam code: quiz04W3rK
(A) : 11 students
(B) : 24 students
(C) : 11 students
(D) : 5 students

Success level: 22%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.43

Physics midterm question: net force of two opposite forces on accelerating object

Physics 205A Midterm 1, fall semester 2018
Cuesta College, San Luis Obispo, CA

A physics question on an online discussion board[*] was asked and answered:

P-dog: If there are only two opposing forces acting on an accelerating object, is it possible for the net force to ever be larger than either of those two forces?
OSG: No.

Discuss why this answer is correct, and how you know this. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.

[*] answers.yahoo.com/question/index?qid=20180916184605AASRFkI.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram, and discusses/demonstrates:
  1. the net force is the vector addition of all forces (in this case, only two forces) acting on the same object; and
  2. since these two forces are opposite in direction, one force must be larger than the other in order for the object to be accelerating (Newton's second law); then
  3. the net force cannot be more than the larger of the two forces acting on the object, as the net force is the difference of the magnitudes of these two forces (i.e., the magnitude of the larger force minus the smaller magnitude force). (It is also possible that the net force could be smaller than both of the two forces acting on the object.)
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's laws to a free-body diagram.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic application of Newton's laws to the forces on a free-body diagram.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01g4iN
p: 33 students
r: 11 students
t: 2 students
v: 10 students
x: 1 student
y: 0 students
z: 1 student

A sample "p" response (from student 1842):
Another sample "p" response (from student 1996):

Physics midterm question: stuck or unstuck crate?

Physics 205A Midterm 1, fall semester 2018
Cuesta College, San Luis Obispo, CA

A force is applied to the right on a 5.0 kg crate on a horizontal floor that has a static friction coefficient µ_s = 0.35. Initially the crate is stationary. Discuss whether or not the crate will remain stationary if the magnitude of the applied force is slowly increased from zero up to a value just below the magnitude of the normal force of the floor on the crate. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram, and discusses/demonstrates:
  1. the crate has two vertical forces acting on it:

     Weight force of Earth on crate (m·g = 49 N, downwards),
     Normal force of table on crate (49 N, upwards),

     and since there is no vertical motion, these two vertical forces must be equal in magnitude due to Newton's first law;
  2. the maximum amount of static friction force that must be overcome in order to unstick the crate is µ_s⋅N = (0.35)·(49 N) = 17 N; and
  3. since the applied force is slowly increased is "slowly increased from zero up to a value just below the magnitude of the normal force of the floor on the crate" (from 0 N up to 49 N), then the applied force will eventually be greater than maximum static friction force, such that the crate will unstick (and begin to accelerate to the right).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Some substantive attempt at analyzing how/if the maximum static friction force is overcome using Newton's laws and a free-body diagram.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at analyzing how/if the maximum static friction force is overcome using Newton's laws and a free-body diagram.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic attempt at using Newton's laws and a free-body diagram.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01g4iN
p: 20 students
r: 7 students
t: 7 students
v: 12 students
x: 11 students
y: 0 students
z: 0 students

A sample "p" response (from student 6577):

Physics midterm question: net forces on car and bus driving over hills

Physics 205A Midterm 1, fall semester 2018
Cuesta College, San Luis Obispo, CA

A 1,800 kg car and a 10,000 kg bus both drive over hills with the same circular radius. Both the car and the bus are still in contact with their hills. Discuss whether the car should drive at a faster or slower speed than the bus in order for them to have the same magnitude net force at the top of their hills. Explain your reasoning by using free-body diagram(s), the properties of forces and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagrams are optional, but primarily discusses/demonstrates:
  1. Newton's second law for uniform circular motion applies, such that for both the car and bus, the net force ΣF (the resultant of the upwards normal force and the downwards weight force) must have a magnitude m⋅v²/r) and point in towards the center of the circular motion--which is vertically downwards; and
  2. since both the car and the bus drive over hills with the same circular radius, then in order for them to have the same magnitude net force, the (less massive) car must drive at a faster speed, while the (more massive) bus must drive at a slower speed.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Some substantive attempt at applying Newton's second law for uniform circular motion.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at applying Newton's first law or third law for uniform circular motion.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic application of Newton's laws.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01g4iN
p: 26 students
r: 7 students
t: 11 students
v: 9 students
x: 3 students
y: 0 students
z: 2 students

A sample "p" response (from student 1408):

Physics quiz question: truck-pushed crate

Physics 205A Quiz 3, fall semester 2018
Cuesta College, San Luis Obispo, CA

A truck applies a force of 1,700 N to the right on a 400 kg crate, such that the crate (with the truck) moves to the right across a road with an acceleration of 
0.50 m/s². The road is not frictionless. Newton's __________ law tells you that these two forces are equal in magnitude and opposite in direction:

Applied force of truck on the crate.
Kinetic friction force of road on the crate.

(A) first.
(B) second.
(C) third.
(D) (These forces are not equal in magnitude and/or opposite in direction.)

Correct answer (highlight to unhide): (D)

The crate has two vertical forces acting on it:

Weight force of Earth on crate (downwards).
Normal force of road on crate (upwards).

Because the crate has no vertical motion, these vertical two forces must be equal in magnitude and opposite in direction, due to Newton's first law, such that these two forces produce a vertical net force of zero.

The crate has two horizontal forces acting on it:

Applied force of truck on the crate (1,700 N to the right).
Kinetic friction force of road on the crate (to the left, against the direction of motion).

Because the crate is accelerating (to the right), then from Newton's second law (as its motion is changing, and not constant) these two horizontal forces do not cancel out, and must sum to a horizontal net force that points in the direction of acceleration. Thus the applied force of truck on the crate is opposite in direction to, but is greater in magnitude than the kinetic friction force of road on the crate.

Sections 70854, 70855
Exam code: quiz03pRH5
(A) : 9 students
(B) : 12 students
(C) : 4 students
(D) : 31 students

Success level: 55%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.60

Physics quiz archive: vectors, projectile motion, forces

Physics 205A Quiz 3, fall semester 2018
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855 version 1
Exam code: quiz03pRH5



Sections 70854, 70855 results

0- 6 :
7-12 :	***** [low = 9]
13-18 :	**********
19-24 :	******************* [mean = 23.1+/- 6.0]
25-30 :	********************** [high = 30]

Online reading assignment: applications of Newton's laws (friction)

Physics 205A, fall semester 2018
Cuesta College, San Luis Obispo, CA

Students have a bi-weekly online reading assignment (hosted by SurveyMonkey.com), where they answer questions based on reading their textbook, material covered in previous lectures, opinion questions, and/or asking (anonymous) questions or making (anonymous) comments. Full credit is given for completing the online reading assignment before next week's lecture, regardless if whether their answers are correct/incorrect. Selected results/questions/comments are addressed by the instructor at the start of the following lecture.

The following questions were asked on reading textbook chapters on applications of Newton's laws (emphasizing static and kinetic friction).


Selected/edited responses are given below.

Describe what you understand from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically demonstrate your level of understanding.

"If you have a constant speed and direction then Newton's first law applies. If you have a change in speed and/or direction then Newton's second law applies."

"Static friction goes against the impending relative motion between two objects and that kinetic friction goes against the relative sliding motion between two surfaces. Newton's second law can be used to find the acceleration of an object."

"I understand the difference between both static friction and kinetic friction. Static friction is the force needed to keep a stationary object at rest, for example when you go rock climbing. The static frictional forces help support someone's weight as they press against the walls of the rock and create large amounts of normal forces. Now once two surfaces begin sliding over one another then kinetic friction is produced. And kinetic friction is what slows down a moving object. For example, when you push a chair across the floor, the initial force to actually move the object requires greater force, but it takes less force to keep the object sliding."

"Static friction depends on the amount of force that is applied to stationary object. Kinetic friction is applied when an object is sliding across the floor."

"The difference between static and kinetic friction. Static friction is when an object is at rest, and once it starts to move it has kinetic friction acting on it."

"How to distinguish between Newton's first law and third law. Newton's first law relates two or more forces acting on the same object (if motion is constant), while Newton's third law relates the same force acting on two different objects."

"The coefficient of static friction is equal to or higher than the coefficient of kinetic friction."

"For the most part I seemed to understand what I read but I had to read it twice."

"I kind of understand the whole concept of friction forces, but it is still very confusing."

"Static friction has a maximum value and once that value is surpassed, then I think of it like the transfer of friction from static to kinetic. It's interesting that the maximum static friction value is independent of surface area, which I did not expect. However, the maximum value of static friction does depend on the type of surface materials."

Describe what you found confusing from the assigned textbook reading or presentation preview. Your description (2-3 sentences) should specifically identify the concept(s) that you do not understand.

"I need some more practice on distinguishing between newtons first and third laws in different situations. I also need to memorize what 'POF-OST-ITO' stands for."

"Static friction force--it does not make sense to me that there can be a frictional force when the object is stationary and why tugging on the object doesn't cause it to move, but I think I understand now."

"Knowing what equations to use to solve for a static/kinetic friction problem."

"I didn't understand why the magnitude of the kinetic frictional force is proportional to the magnitude of the normal force."

"The relationship between friction and normal force."

"Difference between static and kinetic friction."

"What I found to be the most confusing from the assigned reading was the difference between static and kinetic friction. I think that static friction is applied force to keep an object stationary and then kinetic friction is when two things slide over each other, but I am not sure."

"All of the equations that were given in the book, there was just a lot of information thrown out all at once, but I should be able to understand it after I have to use it for a problem."

"Pretty much this whole chapter was confusing. I didn't take as much time reading it as the other chapters though, so that could play a part in that."

"I found it confusing knowing which law to apply in terms of Newton's first law or Newton's third law, but it was better explained in class and I understand now."

What is the meaning of the "normal" in the "normal force?"

"'Normal' means perpendicular to the surface."

"The 'normal' comes from the fact that it is always there when an object is in contact against a surface?"

"Any forces already acting on the object of concern?"

"'Normal' force usually refers to forces that act naturally in a sense?"

"Does it mean 'instantaneous?'"

The SI (Système International) units of the static friction coefficient µ_s and the kinetic friction coefficient µ_k are:

"The coefficients are unitless."

"Trick question, there are none."

"I am truly confused."

Identify the magnitude of the static friction force f_s for each of the following situations of a box that is initially stationary on a horizontal floor. (Only correct responses shown.)

No external horizontal forces applied to it, so it remains stuck to the floor:
f_s = 0. [83%]

An external horizontal force applied to it, but still remains stuck to the floor:
f_s = some value between 0 and µ_s·N. [68%]

An external horizontal force applied to it, at the threshold of nearly becoming unstuck:
f_s = µ_s·N. [71%]

Identify the magnitude of the kinetic friction force f_k for each of the following situations of a box that is already sliding across a horizontal floor. (Only correct responses shown.)

No external horizontal forces applied on it, so it slows down:
f_k = µ_k·N. [22%]

An external horizontal force applied in the forward direction, but not enough to keep the box going so it still gradually slows down:
f_k = µ_k·N. [17%]

An external horizontal force applied in the forward direction, just enough to keep the box going at a constant speed:
f_k = µ_k·N. [39%]

An external horizontal force applied in the forward direction, enough to gradually increase the speed of the box:
f_k = µ_k·N. [37%]

An external horizontal force applied in the backwards direction, such that the box slows down:
f_k = µ_k·N. [22%]

Ask the instructor an anonymous question, or make a comment. Selected questions/comments may be discussed in class.

"I struggled with these questions and probably need some more explanation on them."

"I don't understand how to determine which forces cancel out. Also, if the net force is negative, what would happen?" (Determining which forces are exerted on an object depends on the wording of the problem. Deciding whether these forces cancel or not depends on whether the motion is constant or changing. If a net force is negative, then it points in the negative direction (depending on how you defined your positive direction); but usually we denote magnitudes of forces (keeping them all positive), and explicitly denoting their direction in words (left/right/up/down) or arrows.)

"When would f_k = some value between 0 and µ_k·N? Is that possible?" (That would never be possible. The kinetic friction force (assuming that the object is already unstuck, and has motion (constant or changing) would always be equal to µ_k·N.)

"Could you please go over what the values of static and kinetic frictional forces mean? I'm having a hard time grasping the concept of when kinetic friction force equals zero, and so on." ("Kinetic" means "in motion," so if the object is still stuck to a surface, then kinetic friction force is zero.)

"I could not find the presentation preview." (There were no presentation slides for this topic; just the textbook chapters.)

"You are really helpful going over the chapters and material in class. It helps me out a lot." (You're helping me out a lot by telling me what you specifically understand or are confused about when I prepare for each class.)

Physics midterm question: proposed POF-OSO-WTF test for Newton's first law forces

Physics 205A Midterm 1, fall semester 2017
Cuesta College, San Luis Obispo, CA

A Physics 205A student proposes a three-part "POF-OSO-WTF" test for identifying a pair of forces related by Newton's first law:

__ POF = Pair of Opposite Forces
__ OSO = On the Same Object
__ WTF = With just Two Forces

Determine whether or not this "POF-OSO-WTF" test can correctly identify a pair of forces related by Newton's first law acting on a 600 kg load of cargo that rests on the bed of a stationary truck. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram, and discusses/demonstrates:
  1. that there are two forces acting on the 600 kg cargo load:

     weight force w of Earth on cargo load (downwards),
     normal force N of truck on cargo load (upwards); and

  2. since the cargo load is stationary, then from Newton's first law these two forces are equal in magnitude in order for the net force to be zero; then
  3. tests each part of the presumptive "POF-OSO-WTF" test for the cargo load, where the weight force and the normal force are the "Pair of Opposite Forces," and both act "On the Same Object" (the cargo load), and this test applies to the cargo load "With just Two Forces." (Note that it cannot be applied to the truck, even though Newton's first law applies to it as well, as there are three forces acting on it.)
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May apply the "POF-OST-WTF" test to forces on the truck (when instructed to identify a pair of forces acting on the cargo load), or to an interaction pair of forces (which are actually related by Newton's third law, despite being instructed to identify a pair of forces related by Newton's first law); but at least has (1)-(2) above complete and correct. Or may have (2)-(3) essentially complete/correct, but has minor errors in forces on free body diagram, such as involving forces that do not act on the cargo load, and/or forces exerted by the cargo load.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Essentially only (1)-(2) complete/correct, but does not explicitly discuss relevance of "POF-OSO-WTF" test.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some substantive attempt at applying Newton's laws to a free-body diagram
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic application of Newton's laws to the forces on a free-body diagram.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01mOoL
p: 32 students
r: 10 students
t: 8 students
v: 3 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 5102):

Physics midterm question: comparing accelerations of sliding boxes

Physics 205A Midterm 1, fall semester 2017
Cuesta College, San Luis Obispo, CA

Two 0.5 kg boxes are moving to the left, both slowing down while pushed to the right by an applied force of 10 N. Discuss why the box sliding on a frictionless floor will have a smaller magnitude of acceleration than the box sliding on a floor that has friction. Explain your reasoning using free-body diagram(s), the properties of forces, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagrams, and discusses/demonstrates:
  1. because both boxes are slowing down while moving to the left, Newton's second law applies along the horizontal direction, such that the net force (and acceleration) for each box must point to the right; and
  2. for the box on the frictionless floor, there is only horizontal force acting on the box:

     applied force F_applied on the box (10 N, to the right),

     such that the net force on this box must be 10 N to the right; and
  3. for the box on the floor with friction, there are two horizontal forces acting on the box:

     applied force F_applied on the box (10 N, to the right),
     kinetic friction force f_k of floor on the box (? N, to the right),

     and since these two forces act in the same direction, the net force on this box is the addition of these two force magnitudes (10 N + ? N), and thus must be greater than 10 N;
  4. since the boxes have the same mass, the box with the greater net force magnitude will have the greater acceleration magnitude.
  (May have only implied (1) and (4) above, and/or ignored discussion of vertical forces and Newton's first law, as long as discussion of (2)-(3) is clear and complete.)
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. May have kinetic friction (or other) force vectors incorrectly drawn in opposite directions, additional forces (or other types of vectors) added in; but at least recognizes that for the box on the frictionless floor there is only one force that contributes to the net force, while for the box on the floor with friction there are two forces that contribute to the net force.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some garbled attempt at applying Newton's laws to free-body diagram(s).
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic application of Newton's laws to the forces on free-body diagram(s).
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01mOoL
p: 14 students
r: 19 students
t: 5 students
v: 14 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student):
Another sample "p" response (from student):
Yet nother sample "p" response (from student):

Physics midterm question: forces on string-twirled ball

Physics 205A Midterm 1, fall semester 2017
Cuesta College, San Luis Obispo, CA

A Physics 205A student twirls a ball attached to a string in a vertical circle with constant speed. Discuss why it is possible for the tension force of her hand on the ball to have the same magnitude as the weight force of Earth on the ball, when the ball is swinging through the highest part of the circle. Explain your reasoning by using free-body diagram(s), the properties of forces and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagram, and discusses/demonstrates:
  1. Newton's second law for uniform circular motion applies, such that while the ball is swinging through the highest part of the circle, the net force ΣF (of magnitude m⋅v²/r) must point in towards the center of the circular motion--which is vertically downwards; and
  2. while the ball is swinging through the highest part of the circle, there are two forces acting on the ball:

     weight force w of Earth on the ball (m⋅g, downwards);
     tension force T of hand on the ball (downwards),

  3. since both these two downwards forces add together to get the downwards net force, it is possible that they each have the same magnitude of one-half of m⋅v²/r (for certain values of mass m, speed v, and radius r).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Two of the three points (1)-(3) correct/complete.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. May argue T (upwards) = w (downwards) because of Newton's first law; or may draw T and w as both acting downwards but does not explicitly relate how Newton's second law requires a downwards net force that is met by their addition, and/or introduces normal forces, Newton's third law, etc.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some substantive attempt at applying Newton's laws to free-body diagram.
• x:
  Implementation of ideas, but credit given for effort rather than merit. No systematic application of Newton's laws to the forces on free-body diagram.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm01mOoL
p: 10 students
r: 7 students
t: 34 students
v: 11 students
x: 2 students
y: 0 students
z: 0 students

A sample "p" response (from student 4632):
Another sample "p" response (from student 8659):
A sample "t" response (from student 2500):

Physics quiz question: horizontal forces on book sliding at constant speed

Physics 205A Quiz 3, fall semester 2017
Cuesta College, San Luis Obispo, CA

A force of 20 N is applied to the right on a 1.5 kg book such that it continues to slide to the right with constant speed. The table is not frictionless. Newton's __________ law tells you that these two forces are equal in magnitude and opposite in direction:

Applied force on the book.
Kinetic friction force of the table on the book.

(A) first.
(B) second.
(C) third.
(D) (These forces are not equal in magnitude and/or opposite in direction.)

Correct answer (highlight to unhide): (A)

The book has two vertical forces acting on it:

Weight force of Earth on book (downwards).
Normal force of book on suitcase (upwards).

Because the book has no vertical motion, these vertical two forces are equal in magnitude and opposite in direction, due to Newton's first law, such that these two forces produce a vertical net force of zero.

The book has two horizontal forces acting on it:

Kinetic friction force of floor on book (to the left, against the direction of motion).
External applied force on book (20 N, to the right).

Because the book is moving at constant speed (to the right), then its horizontal acceleration is zero, and by Newton's first law these two horizontal forces are equal in magnitude and opposite in direction in order to produce a horizontal net force of zero.

Sections 70854, 70855
Exam code: quiz03T4uC
(A) : 16 students
(B) : 14 students
(C) : 8 students
(D) : 16 students

Success level: 30%
Discrimination index (Aubrecht & Aubrecht, 1983): 0

Physics quiz question: comparing directions and magnitudes of forces

Physics 205A Quiz 3, fall semester 2017
Cuesta College, San Luis Obispo, CA

"PACCAR Financial - Watkins Trucking 1"
TruckPR
flic.kr/p/oMJDhg

A 600 kg load of cargo rests on the bed of a 8,850 kg stationary truck[*]. Newton's __________ law tells you that these two forces are equal in magnitude and opposite in direction:

Normal force of ground on the truck.
Normal force of truck on the cargo load.

(A) first.
(B) second.
(C) third.
(D) (These forces are not equal in magnitude and/or opposite in direction.)

[*] trucktrend.com/truck-reviews/163-0501-gm-kodiak-topkick-4x4/.

Correct answer (highlight to unhide): (D)

The cargo load has two forces acting on it:

Weight force of Earth on the cargo load (m_cargo·g = 5.9×10³ N, downwards).
Normal force of truck on the cargo load (upwards).

Because the cargo load is stationary, these two forces are equal in magnitude (each 5.9×10³ N) and opposite in direction, due to Newton's first law, so the upwards normal force of truck on the cargo load must be 5.9×10³ N in magnitude.

The truck has three forces acting on it:

Weight force of Earth on the truck (m_truck·g = 8.67×10⁴ N, downwards).
Normal force of cargo load on the truck (downwards).
Normal force of ground on the truck (upwards).

Because the direction of the normal force of ground on the truck is already known to be upwards, and the direction of the normal force of truck on the cargo load is also known to be upwards, then they cannot be forces that are opposite in direction.

(To determine magnitudes of the two forces in question, because the truck is stationary, the magnitudes of the two downward forces added together must equal the magnitude of the upwards force, due to Newton's first law. From Newton's third law, the downwards normal force of the cargo load on the truck must be equal in magnitude and opposite in direction to the upwards normal force of the truck on the cargo load, so they are each 5.9×10³ N.)

(Subsequently, the upwards normal force of the ground on the truck will be equal to the addition of the downwards force of the weight force of Earth on the truck (8.67×10⁴ N) with the downwards normal force of the cargo load on the truck (5.9×10³ N), and thus is 9.26×10⁴ N.)

(Then since the magnitude of the normal force of ground on the truck is 9.26×10⁴ N, and the magnitude of the normal force of truck on the cargo load is 5.9×10³ N, then they cannot be forces that have the same magnitude, as well.)

Sections 70854, 70855
Exam code: quiz03T4uC
(A) : 8 students
(B) : 2 students
(C) : 10 students
(D) : 34 students

Success level: 63%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.81

Physics quiz question: force of cargo load on truck magnitude

Physics 205A Quiz 3, fall semester 2017
Cuesta College, San Luis Obispo, CA

"PACCAR Financial - Watkins Trucking 1"
TruckPR
flic.kr/p/oMJDhg

A 600 kg load of cargo rests on the bed of a 8,850 kg stationary truck[*]. The magnitude of the normal force of the cargo load on the truck is:
(A) 0 N.
(B) 5.9×10³ N.
(C) 8.67×10⁴ N.
(D) 9.26×10⁴ N.

[*] trucktrend.com/truck-reviews/163-0501-gm-kodiak-topkick-4x4/.

Correct answer (highlight to unhide): (B)

The cargo load has two forces acting on it:

Weight force of Earth on the cargo load (m_cargo·g = 5.9×10³ N, downwards).
Normal force of truck on the cargo load (upwards).

Because the cargo load is stationary, these two forces are equal in magnitude (each 5.9×10³ N) and opposite in direction, due to Newton's first law, so the upwards normal force of truck on the cargo load must be 5.9×10³ N in magnitude.

The truck has three forces acting on it:

Weight force of Earth on the truck (m_truck·g = 8.67×10⁴ N, downwards).
Normal force of cargo load on the truck (downwards).
Normal force of ground on the truck (upwards).

Because the truck is stationary, the magnitudes of the two downward forces added together must equal the magnitude of the upwards force, due to Newton's first law. From Newton's third law, the downwards normal force of the cargo load on the truck must be equal in magnitude and opposite in direction to the upwards normal force of the truck on the cargo load, so they are each 5.9×10³ N.

(Subsequently, the upwards normal force of the ground on the truck will be equal to the addition of the downwards force of the weight force of Earth on the truck (8.67×10⁴ N) with the downwards normal force of the cargo load on the truck (5.9×10³ N), and thus is 9.26×10⁴ N.)

Sections 70854, 70855
Exam code: quiz03T4uC
(A) : 10 students
(B) : 35 students
(C) : 5 students
(D) : 4 students

Success level: 65%
Discrimination index (Aubrecht & Aubrecht, 1983): –0.06

Physics quiz archive: vectors, projectile motion, forces

Physics 205A Quiz 3, fall semester 2017
Cuesta College, San Luis Obispo, CA
Sections 70854, 70855 version 1
Exam code: quiz03T4uC



Sections 70854, 70855 results

0- 6 :
7-12 :	********* [low = 9]
13-18 :	****************
19-24 :	*********************** [mean = 19.2 +/- 5.2]
25-30 :	****** [high = 27]