Physics midterm question: comparing vertical forces supporting tilted beams

Physics 205A Midterm 2, fall semester 2019
Cuesta College, San Luis Obispo, CA

A force F₁ pulls up at the end of a uniform beam to hold it stationary at an angle of 80° above the horizontal, and a force F₂ pulls up at the end of an identical uniform beam to hold it stationary at an angle of 10° above the horizontal. (Calculate all torques with respect to the pivot, located at the base of the beams.) Discuss why these forces 
 F₁ and F₂ have the same magnitude. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. the magnitude of the weight force w is the same for both higher and lower beams; and
  2. for each beam, the lever arm for the applied force F is twice the lever for the weight force w (2⋅ℓ_w = ℓ_F); and
  3. Newton's first law for rotations applies to both higher and lower beams, where the ccw force torque F⋅(ℓ_F) and cw weight torque w⋅(ℓ_w) must balance each other out, and so: F⋅(ℓ_F) = w⋅(ℓ_w), F = w⋅(ℓ_w/ℓ_F) = w⋅(ℓ_w/(2⋅ℓ_w)) = w/2; such that
  4. the applied forces on the higher and lower beam must be equal in magnitude, as they are both equal to one-half of the weight of the beam.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Does not explicitly note that the ℓ_F lever arm is always twice the ℓ_w lever arm for both situations. Instead, argues that since the ℓ_F and ℓ_w values for the higher beam are both bigger than the respective ℓ_F and ℓ_w values for the lower beam, then the higher beam F = w⋅(ℓ_w/ℓ_F) = w⋅(bigger/bigger) must be equal to the lower beam F = w⋅(ℓ_w/ℓ_F) = w⋅(smaller/smaller), but only implicitly demonstrates how the "bigger/bigger" ratio is exactly equal to the "smaller/smaller" ratio by use of a scaled drawing instead of using geometry/trigonometry, etc.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. As (r), but does not clearly/correctly show ℓ_F and ℓ_w lever arms for both situations. At least has two sets of Newton's first law for rotations, one for the higher beam and one for the the lower beam, setting the ccw torques equal to the cw torques.
• 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 first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02sQm5
p: 3 students
r: 18 students
t: 14 students
v: 14 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 5281):

Physics quiz question: comparing torques on supported square

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA

A uniform 0.40 m × 0.40 m square with 
a mass of 0.50 kg is pivoted at one corner. It is supported by a 45° diagonal force at the opposite corner such that the bottom edge is parallel to the ground. Calculate all torques with respect to the corner pivot. The torque exerted by the __________ has a greater magnitude. (A) weight force.
(B) support force.
(C) (There is a tie.)
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (C)

Since the square is in static equilibrium ("is supported") and does not rotate, then the net torque on it is equal to zero (Στ = 0), such that the counterclockwise torque of the weight force on the square must equal the clockwise torque of the support force on the square:

(ccw) τ_w = (cw) τ_Fsupport.

Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 23 students
(B) : 21 students
(C) : 7 students
(D) : 0 students

Success level: 13%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.24

Physics quiz question: weight torque on supported square

Physics 205A Quiz 5, fall semester 2019
Cuesta College, San Luis Obispo, CA

A uniform 0.40 m × 0.40 m square with 
a mass of 0.50 kg is pivoted at one corner. It is supported by a 45° diagonal force at the opposite corner such that the bottom edge is parallel to the ground. Calculate all torques with respect to the corner pivot. The magnitude of the torque exerted by the weight force is:
(A) 0.98 N·m.
(B) 1.4 N·m.
(C) 2.0 N·m.
(D) 2.8 N·m.

Correct answer (highlight to unhide): (A)

The weight force w of Earth acting on the beam acts at the center of gravity, directly downwards with a magnitude m·g. The perpendicular lever arm ℓ for the weight force on the beam must extend from the pivot point to perpendicularly intercept the weight force line of action (which lies along the weight force vector), such that this will be a horizontal line that is half the side of the square:

ℓ = 0.20 m.

The magnitude of the (counterclockwise) torque exerted by the weight force on the beam is then:

τ = w·ℓ,

τ = (m·g)·(0.20 m),

τ = ((0.50 kg)·(9.80 m/s²))·(0.20 m),

τ = 0.98 N·m.

(Response (B) is m·g·(0.20 m)/sin(45°); response (C) is m·g·(0.40 m); and response (D) is m·g·(0.40 m)/sin(45°).)

Sections 70854, 70855
Exam code: quiz05Gu1L
(A) : 26 students
(B) : 16 students
(C) : 6 students
(D) : 4 students

Success level: 50%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.65

Physics midterm question: comparing horizontal forces supporting tilted beams

Physics 205A Midterm 2, fall semester 2018
Cuesta College, San Luis Obispo, CA

A horizontal force is applied to hold a uniform beam stationary at an angle of 80° above the horizontal, and another horizontal force is applied to hold it stationary at an angle of 10° above the horizontal. (Calculate all torques with respect to the pivot, located at the base of the beam.) Discuss why less force required to hold the beam when it is at the higher 80° angle. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. the magnitude of the weight force w is the same for both higher and lower beams; but
  2. the lever arms ℓ_F are not the same, where ℓ_F is longer for the higher beam, and shorter for the lower beam; and
  3. the lever arms ℓ_w are not the same, where ℓ_w is shorter for the higher beam, and longer for the lower beam; and
  4. Newton's first law for rotations applies to both higher and lower beams, where the ccw force τ = F⋅ℓ_F and cw weight τ = w⋅ℓ_w must balance each other out, and so:

     F⋅ℓ_F = w⋅ℓ_w,
     
     F = w⋅(ℓ_w/ℓ_F);

     such that
  5. for the higher beam, the shorter ℓ_w in the numerator and longer ℓ_F in the denominator means that the applied force F is smaller than for the lower beam (where it has a longer ℓ_w in the numerator and a shorter ℓ_F in the denominator).
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. At least demonstrates that for the higher beam ℓ_w is shorter and ℓ_F is longer, but typically discusses only how one of these contributes to making the applied force smaller.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Typically argues that F is smaller for the higher beam because ℓ_F is larger (while claiming ℓ_w is the same for both beams); or F is smaller for the higher beam because ℓ_w is smaller (while claiming ℓ_F is the same for both beams).
• 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 first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02r3iN
p: 17 students
r: 4 students
t: 21 students
v: 10 students
x: 5 students
y: 0 students
z: 0 students

A sample "p" response (from student 5250):

Physics quiz question: falling beam torque

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

A uniform beam with a 0.15 kg mass and 1.2 m length pivots at one end as it falls over starting from an angle of 5° from the vertical. Calculate all torques with respect to the bottom of the beam. While at its starting point of 5°, the magnitude of the torque exerted by the weight force on the beam is:
(A) 0.077 N·m.
(B) 0.88 N·m.
(C) 1.5 N·m.
(D) 1.8 N·m.

Correct answer (highlight to unhide): (A)

The weight force w of Earth acting on the beam acts at the center of gravity, directly downwards with a magnitude m·g. The perpendicular lever arm ℓ for the weight force on the beam must extend from the pivot point to perpendicularly intercept the weight force line of action (which lies along the weight force vector), such that this will be a horizontal line of length:

ℓ = (L/2)·cos(85°).

The magnitude of the (clockwise) torque exerted by the weight force on the beam is then:

τ = w·ℓ,

τ = (m·g)·(L/2)·cos(85°),

τ = ((0.15 kg)·(9.80 m/s²))·((1.2 m)/2)·cos(85°),

τ = 0.0768713651... N·m,

or to two significant figures, the torque of the weight force on the beam has a magnitude of 0.077 N·m.

(Response (B) is (m·g)·(L/2)·sin(85°); response (C) is m·g; and response (D) is m·g·L.)

Sections 70854, 70855
Exam code: quiz05Ro74
(A) : 6 students
(B) : 11 students
(C) : 4 students
(D) : 31 students

Success level: 12%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.25

Physics midterm question: tension in cable lowering a boom crane

Physics 205A Midterm 2, fall semester 2017
Cuesta College, San Luis Obispo, CA

A cable anchored to a wall suspends a uniform beam, which can either be held 40° above the horizontal, or 40° below the horizontal. (Calculate all torques with respect to the pivot, located at the base of the beam.) Discuss why there is less tension in the cable when the beam is held 40° above the horizontal, compared to when it is held 40° below the horizontal. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. for either case (40° above or below the horizontal), the ccw cable tension torque = ℓ_T⋅T and cw weight torque = ℓ_w⋅w; and
  2. since Newton's first law for rotations applies to both cases, the ccw cable tension torque equals the cw weight torque such that ℓ_T⋅T = ℓ_w⋅w; and
  3. the perpendicular lever arm ℓ_w and the weight force is the same for both cases, where ℓ_w = (L/2)·cos40° and w = m⋅g; and
  4. since the cw weight torque is the same for both cases, the ccw cable tension torque is also the same for both cases; and
  5. the first case (40° above the horizontal) has a larger perpendicular lever arm ℓ_T, such that its tension T has a smaller magnitude compared to the second case (40° below the horizontal), with a smaller perpendicular lever arm ℓ_T, such that its tension T has a greater magnitude. Must at least clearly draw these ℓ_T lever arms to compare their relative lengths.
• 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 incorrect or missing cable lever arm.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Has at least three of the (1)-(5) steps above.
• 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 first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02bu2Z
p: 14 students
r: 3 students
t: 20 students
v: 10 students
x: 3 students
y: 0 students
z: 0 students

A sample "p" response (from student 2635):

Physics quiz question: comparing lever arms for loaded beam forces

Physics 205A Quiz 5, fall semester 2017
Cuesta College, San Luis Obispo, CA

A uniform beam with a mass of 10 kg and length 2.0 m has a 3.0 kg load hanging from its end, and is suspended by a horizontal cable attached to a wall. (Calculate all torques with respect to the pivot.) The __________ force has the longest perpendicular lever arm.
(A) horizontal cable tension.
(B) beam weight.
(C) load.
(D) (There is a tie.)

Correct answer (highlight to unhide): (A)

The weight of the boom acts at its center of gravity, straight downwards. The perpendicular lever arm ℓ_w for the weight force w extends from the pivot to perpendicularly intercept the weight force line of action, such that this will be a horizontal line of length:

ℓ_w = (1.0 m)·cos(70°) = 0.34 m.

Similarly the perpendicular lever arm ℓ_load for the load force F_load extends from the pivot to the perpendicularly intercept the load force of action, such that:

ℓ_load = (2.0 m)·cos(70°) = 0.68 m.

The perpendicular lever arm for the horizontal cable force extends from the pivot point to perpendicularly intercept the tension force of action, such that:

ℓ_cable = (2.0 m)·sin(70°) = 1.9 m.

Sections 70854, 70855
Exam code: quiz05nWaW
(A) : 21 students
(B) : 11 students
(C) : 14 students
(D) : 3 students

Success level: 43%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.53

Physics final exam question: comparing boom crane piston forces

Physics 205A Final Exam, fall semester 2016
Cuesta College, San Luis Obispo, CA

A telescoping boom crane can either be supported by two types of diagonal pistons, attached to the same point along the boom. (Calculate all torques with respect to the pivot, located at the base of the boom, approximated here as a uniform beam.) Determine which piston is exerting a greater magnitude force on the boom, or if there is a tie. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. for either case (piston (A) or piston (B)), the cw piston torque = l_piston⋅F_piston and ccw weight torque = l_weight⋅w; and
  2. since Newton's first law applies to both cases (piston (A) or piston (B)), the ccw torque of the piston on the boom equals the cw torque of the weight on the boom such that l_piston⋅F_piston = l_weight⋅w; then
  3. for piston (A) and piston (B), the ccw torque of the piston on the boom l_piston⋅F_piston has the same value;
  4. the perpendicular lever arm l_piston is smaller for piston (A) and larger for piston (B), and since l_piston⋅F_piston remains constant, then F_piston must be larger for piston (A) and smaller for piston (B).
• 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 only implied Newton's first law in order to compare (equal) piston torques.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. May not have recognized difference in l_piston lever arms (claiming that they are the same, and thus F_piston forces are the same); or recognizes difference in l_piston, but claims that piston (B) has the greater F_piston force.
• 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 first law to torques, forces, and perpendicular lever arms. May have pistons exerting different amounts of torques.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finali0w4
p: 22 students
r: 3 students
t: 19 students
v: 5 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 6436):

Physics midterm question: extended boom crane torques, forces

Physics 205A Midterm 2, fall semester 2016
Cuesta College, San Luis Obispo, CA

A telescoping boom crane extends its length (while keeping the mass of the boom constant), supported by a vertical piston. (Calculate all torques with respect to the pivot, located at the base of the boom, approximated here as a uniform beam.) Discuss why the force exerted by the piston increases as the boom is extended. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. for either case (short boom or long boom), the cw piston torque = ℓ_piston⋅F_piston and ccw weight torque = ℓ_w⋅w; and
  2. since Newton's first law applies to both cases (short boom and long boom), the ccw torque of the piston on the boom equals the cw torque of the weight on the boom such that ℓ_piston⋅F_piston = ℓ_w⋅w; then
  3. for the long boom case compared to the short boom case, the perpendicular lever arm ℓ_w is longer (while w remains the same), and since ℓ_piston remains the same, then F_piston must increase.
• 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 first law to torques, forces, and perpendicular lever arms.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm02oPt0
p: 17 students
r: 10 students
t: 11 students
v: 17 students
x: 1 student
y: 0 students
z: 0 students

A sample "p" response (from student):

Physics quiz question: piston force on boom crane

Physics 205A Quiz 5, fall semester 2016
Cuesta College, San Luis Obispo, CA

A boom crane is supported by a vertical piston. (Calculate all torques with respect to the pivot, located at the base of the boom, approximated here as a uniform beam.) The magnitude of the force applied by the piston is _________ the magnitude of the weight of the boom.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (C)

The weight of the boom acts at its center of gravity, straight downwards. The perpendicular lever arm for the weight force w extends from the pivot to perpendicularly intercept the weight force line of action, such that this will be a horizontal line of length ℓ_w.

The piston force acts straight upwards, and the perpendicular lever arm for the piston force F_piston also extends from the pivot to perpendicularly intercept the piston force line of action, such that this will be also be a horizontal line of slightly shorter length ℓ_piston.

Since the boom is in static equilibrium ("is supported") and does not rotate, then the net torque on it is equal to zero (Στ = 0), such that the clockwise torque of the weight force on the boom must equal the counterclockwise torque of the piston force on the boom:

(cw) τ_w = (ccw) τ_piston.

Substituting in the perpendicular lever arms for these torques:

w·ℓ_w = F_piston·ℓ_piston,

and since ℓ_w > ℓ_piston, then w < F_piston, and thus the force applied by the piston is greater than the magnitude of the weight of the boom.

Sections 70854, 70855, 73320
Exam code: quiz05b0oM
(A) : 7 students
(B) : 24 students
(C) : 24 students
(D) : 0 students

Success level: 42%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.74

Physics quiz question: tension force of string on suspended beam

Physics 205A Quiz 5, fall semester 2015
Cuesta College, San Luis Obispo, CA

A uniform beam (0.60 m length) is suspended by a string that pulls vertically upwards, exerting a torque of magnitude 1.1 N·m. (Calculate all torques with respect to the pivot, located at the other end.) The tension in the string is:
(A) 3.7 N.
(B) 0.92 N.
(C) 0.66 N.
(D) 1.8 N.

Correct answer (highlight to unhide): (A)

The tension force T of the string acts at the top of the beam, along the string. The perpendicular lever arm ℓ for the string tension force must extend from the pivot to perpendicularly intercept the tension force line of action (which lies along the string itself), such that this will be a horizontal line of length:

ℓ = L·cos(60.0°).

The (clockwise) torque exerted by the string on the beam is then:

τ = T·ℓ,

such that the magnitude of the tension force T can be solved for:

T = τ/ℓ/ = τ/(L·cos(60.0°)),

T = (1.1 N·m)/((0.60 m)·cos(60.0°)) = 3.66666666... N,

or to two significant figures, the tension in the string is 3.7 N.

(Response (B) is τ/(2·L); response (C) is τ/2; and response (D) is τ/L.)

Sections 70854, 70855, 73320
Exam code: quiz05bL8D
(A) : 14 students
(B) : 2 students
(C) : 19 students
(D) : 30 students

Success level: 45%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.47

Physics midterm question: tilting up a plywood board

Physics 205A Midterm 2, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Multiple-Choice Question 8.10, Problem 8.19

A Physics 205A student lifts a plywood board and keeps it stationary, while it is just off of the floor, or when it is nearly vertical. In both cases the student exerts a force perpendicular to the top edge of the board, which can be approximated as a uniform beam pivoted at the bottom edge. Discuss why more force must be exerted to hold it just off the floor, compared to when it is nearly vertical. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Complete free-body diagrams with forces and perpendicular lever arms, and discusses/demonstrates:
  1. Newton's first law applies to each case, such that the cw torque of the student equals the ccw torque of the weight;
  2. the weight force is the same in either case, as is the perpendicular lever arm for the student's force;
  3. since the perpendicular lever arm for weight is less for the second case, then the magnitude of the student's force will be less.
• r:
  Nearly correct, but includes minor math errors. As (p), but typically does not explicitly note that the perpendicular lever arm for the student's force is the same in either case (board length L).
• t:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. Typically has a different perpendicular lever arm for the student's force in either case.
• v:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some garbled attempt at applying Newton's first law to torques, forces, and perpendicular lever arms. May apply Newton's first law to equate weight or student torque in one case to the weight or student torque in the other case.
• x:
  Implementation of ideas, but credit given for effort rather than merit. Approach other than that of applying Newton's first law to torques, forces, and perpendicular lever arms.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.

Grading distribution:
Sections 70854, 70855, 73320
Exam code: midterm02veR1
p: 22 students
r: 9 students
t: 13 students
v: 14 students
x: 8 students
y: 0 students
z: 0 students

A sample "p" response (from student 9950):
A sample "t" response (from student 1991), with the student's lever arm being different for either case, while the torque produced by the weight force remains the same:

Physics quiz question: tilted Murphy bed

Physics 205A Quiz 5, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 8.19

"P-Town_0022"
David Boyle
flic.kr/p/eXzCF

A women pushes perpendicularly at the top edge of a fold-down Murphy bed frame (2.1 m length, 91 kg mass) to keep it stationary. Approximate the bed frame as a uniform beam raised at an angle of 55° from the horizontal. (Calculate all torques with respect to the pivot, located at the floor edge.) The magnitude of the force applied by the woman is _________ the magnitude of the weight of the bed frame.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (A)

The force of the woman is already acting perpendicular to the bed, so the perpendicular lever arm ℓ_woman is simply L = 2.1 m.

The weight of the bed acts at its center of gravity, straight downwards. The perpendicular lever arm for the weight force extends from the pivot to perpendicularly intercept the weight force line of action, such that this will be a horizontal line of length:

ℓ_w = (L/2)·cos(55°).

Since the bed is in static equilibrium and does not rotate, then the net torque on it is also equal to zero (Στ = 0), such that the clockwise torque of the woman on the bed must equal the counterclockwise torque of weight acting on the bed:

(cw) τ_woman = (ccw) τ_w.

Substituting in the perpendicular lever arms for these torques:

F_woman·ℓ_woman = w·ℓ_w,

F_woman·(L) = w·((L/2)·cos(55°)),

F_post = w·(1/2)·cos(25°),

F_post = w·(0.29).

Because the force of the woman on the bed has a longer perpendicular lever arm than the perpendicular lever arm of the weight on the bed, then the force of the woman on the beam has a smaller magnitude than the weight of the bed.

Sections 70854, 70855, 73320
Exam code: quiz05mRp4
(A) : 25 students
(B) : 22 students
(C) : 17 students
(D) : 0 students

Success level: 39%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.79

Physics quiz question: diagonally-propped beam

Physics 205A Quiz 5, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 8.19, 8.35

A uniform beam of length 2.7 m is mounted on a pivot at one end, and supported at its center by a post perpendicular to the beam. (Calculate all torques with respect to the pivot.) The magnitude of the force of the post on the beam is __________ the magnitude of the weight of the beam.
(A) less than.
(B) equal to.
(C) greater than.
(D) (Not enough information is given.)

Correct answer (highlight to unhide): (A)

The post force on the beam is already acting perpendicular to the beam, so its perpendicular lever arm ℓ_post is simply L/2.

The weight of the beam acts at its center of gravity, straight downwards. The perpendicular lever arm for the weight force extends from the pivot to perpendicularly intercept the weight force line of action, such that this will be a horizontal line of length:

ℓ_w = (L/2)·cos25°.

Since the beam is (implied to be) in static equilibrium and does not rotate, then the net torque on it is also equal to zero (Στ = 0), such that the counterclockwise torque of the post on the beam must equal the clockwise torque of weight acting on the beam:

τ_post = τ_w.

Substituting in the perpendicular lever arms for these torques:

F_post·ℓ_post = w·ℓ_w,

F_post·(L/2) = w·((L/2)·cos25°),

F_post = w·cos25°,

F_post = w·(0.91).

Because the force of the post on the beam has a longer perpendicular lever arm than the perpendicular lever arm of the weight on the beam, then the force of the post on the beam has a smaller magnitude than the weight of the beam.

Sections 70854, 70855
Exam code: quiz05t0rQ
(A) : 15 students
(B) : 25 students
(C) : 10 students
(D) : 0 students

Success level: 29%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.34

Physics final exam question: climber rappelling down cliff edge

Physics 205A Final Exam, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 8.19

"Transportation Soldiers conduct rappel tower training"
Virginia Guard Public Affairs
flic.kr/p/cQRGch

A climber is approximated as a uniform beam, with a pivot point at the feet, and a rope attached to the center. The climber is about to rappel back down from a cliff edge, leaning back with the rope initially horizontal. A little later, the climber is lower, leaning back with the same angle, but the rope is no longer horizontal. Discuss why there is less tension in the rope for the latter case. (Ignore stretching in the rope.) Explain your reasoning using diagram(s) with locations of forces and lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. Understands that (1) the counterclockwise torque due to weight is the same in either case, such that from the static equilibrium condition (Newton's first law for rotations), the clockwise torque due to the rope's tension force must be the same in either case, and (2) the larger lever arm r_⊥ for the second case will mean a smaller tension force.
• r:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. Does not explicitly explain in (1) why the clockwise torques due to the rope tension forces are the same.
• t:
  Nearly correct, but argument has conceptual errors, or is incomplete. Discusses that r_⊥ would be larger for the second case, but does not explicitly explain this on a diagram and/or with trigonometry.
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. Some attempt at manipulating the relationship between torque, force, and lever arm.
• x:
  Implementation/application of ideas, but credit given for effort rather than merit. Approach does not substantively use relationship between torque, force, and lever arm.
• y:
  Irrelevant discussion/effectively blank.
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855, 73320
Exam code: finaln0M3
p: 6 students
r: 14 students
t: 19 students
v: 17 students
x: 2 students
y: 1 student
z: 2 students

A sample "p" response (from student 2419):

Physics midterm problem: Russky Bridge scale model

Physics 205A Midterm 2, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 8.35(a), 8.37

[20 points.] A Physics 205A student makes a scale model of the Russky Bridge in Vladivostok, Russia(*), during its construction. A meter stick(**) of uniform density and mass 0.178 kg is attached to a pivot point, and is suspended by two parallel strings that each have the same magnitude tension force. A 0.080 kg mass is attached to the far end of the meter stick. Determine the magnitude of the tension force in either of the strings. Show your work and explain your reasoning.

Image source: Vitaliy Ankov, "Lifting a double bridge bay section during the construction," http://russiaprofile.org/photos/52678_6.html

(*) "The world's largest cable-stayed bridge," http://en.wikipedia.org/wiki/Russky_Bridge.
(**) http://flic.kr/p/duoug.

Solution and grading rubric:

• p = 20/20:
  Correct. Applies Newton's first law (rotational equilibrium) by balancing out the sum of the two clockwise torques (of meter stick's weight, and of the mass hanging at end) with the sum of the two counterclockwise torques (of the middle and end strings, which have the same magnitude tension, but different lever arms), and solves for their tensions, which is 1.2 N. (May have used masses instead of weights for the meter stick and hanging mass in calculating torques.)
• r = 16/20:
  Nearly correct, but includes minor math errors. Effectively solves for tension in a system supported by only one string (either attached to the middle, or the end). Still has correct sum of clockwise torques, and identifies correct lever arm for the relevant string.
• t = 12/20:
  Nearly correct, but approach has conceptual errors, and/or major/compounded math errors. May have either (a) separately set the inner string's torque equal to the torque of the meter stick's weight, and the outer string's torque equal to the torque of the hanging mass' weight, and separately solved for the two different string tensions. Or (b) may have claimed that the two strings exert equal amounts of (counterclockwise) torque, and then solved for at least one of the string tensions. At least identifies four different torques acting on the meter stick, calculates each torque with the proper lever arms, and attempts to apply the rotational equilibrium condition.
• v = 8/20:
  Implementation of right ideas, but in an inconsistent, incomplete, or unorganized manner. Some attempt at finding lever arms and applying rotational equilibrium condition to torques.
• x = 4/20:
  Implementation of ideas, but credit given for effort rather than merit. May involve periods of simple harmonic motion systems.
• y = 2/20:
  Irrelevant discussion/effectively blank.
• z = 0/20:
  Blank.

Grading distribution:
Sections 70854, 70855
Exam code: midterm02gL0u
p: 9 students
r: 6 students
t: 5 students
v: 25 students
x: 9 students
y: 0 students
z: 0 students

A sample "p" response (from student 1223):

Physics quiz question: cable-supported beam

Physics 205A Quiz 5, fall semester 2012
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 8.19, 8.35

A 12 kg uniform beam of length 0.80 m is mounted on a pivot at one end, and supported by a cable attached to the other end. With respect to the pivot, the perpendicular lever arm ℓ for the tension force of the cable on the beam is:
(A) 0.40 m.
(B) 0.46 m.
(C) 0.56 m.
(D) 0.66 m.

Correct answer (highlight to unhide): (B)

The tension force of the cable acts on the beam at the left end, along the cable. The ℓ lever arm for the cable tension force must extend from the pivot to perpendicularly intercept the tension force line of action (which is along the cable itself), such that this will be a diagonal line of length:

ℓ = L·sin35°.

(Response (A) is L/2; response (C) is L·tan35°; response (D) is L·cos35°.)

Sections 70854, 70855
Exam code: quiz05L4mN
(A) : 3 students
(B) : 28 students
(C) : 10 students
(D) : 11 students

Success level: 54%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.42

Online reading assignment: vector operations, projectile motion

Physics 205A, fall semester 2012
Cuesta College, San Luis Obispo, CA

Students have a 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 and previewing a presentation on vector operations and projectile motion.

Selected/edited responses are given below.

Describe something you found interesting from the assigned textbook reading or presentation preview, and explain why this was personally interesting for you.

"Horizontal and vertical motion of something are independent of each other. It would seem that they directly effect each other and that their would some mathematical relation between the two, that they directly or inversely predict each other."

"The experiment with the ball falling straight down and one on arch being at same heights. I was surprised that the ball falling straight didn't reach the floor faster."

"Without air resistance horizontal acceleration does not affect the acceleration of gravity."

Describe something you found confusing from the assigned textbook reading or presentation preview, and explain why this was personally confusing for you.

"I'm kinda sketchy on how to subtract vectors. Adding makes sense but subtracting is kinda vague."

"Why is horizontal velocity constant? And why is horizontal displacement equation only a regular velocity equation, inverted?"

Explain what will happen to the magnitude and/or direction of a vector that is multiplied by -1.

"Direction is flipped, or opposite, but magnitude remains the same. Uh-huh."

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

"It's been a while for trig functions, can we go over the basics? :)"

"Is there a sheet that we can print out with all the formulas that we have been using so far and the 'chain of pain?'" (Links to the constant acceleration motion equations and the "chain of pain" are found on the course website.)

"Please work out problem from start to finish instead of telling us what happens." (Come to posted office hours or make an appointment; ask questions just before/after lecture or via e-mail.)

"Do we ever get to play with gravity on a large scale in this class?" (Well, with small objects in laboratory this week.)

Physics midterm question: cable-supported beam

Physics 205A Midterm 2, fall semester 2011
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problems 8.19, 8.35

A uniform beam is mounted on a pivot at one end, and supported at its center by a cable perpendicular to the beam. Discuss why the magnitude of the tension in the cable is less than the weight of the beam. Explain your reasoning using diagram(s) with locations of forces and perpendicular lever arms, the properties of torques, and Newton's laws.

Solution and grading rubric:

• p:
  Correct. The counterclockwise torque from cable equals the clockwise torque from weight, but the cable has a larger lever arm (L/2) than weight (L·cos(25°)/2), such that the cable tension force must be smaller than the weight force.
  
• 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. Applies Newton's first law for rotations, but r_⊥ discussion is garbled or missing.
  
• v:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. At least attempts to discuss torques, forces, lever arms, but Newton's first law for rotations is not clearly used. May instead use Newton's first law for vertical forces and involve the force of the pivot point, but does not clearly break up tension into vertical and horizontal components.
  
• x:
  Implementation of ideas, but credit given for effort rather than merit. Does not discuss torques, forces, lever arms.
• y:
  Irrelevant discussion/effectively blank.
  
• z:
  Blank.
  

Grading distribution:
Sections 70854, 70855
Exam code: midterm02fR3q
p: 16 students
r: 3 students
t: 10 students
v: 24 students
x: 0 students
y: 0 students
z: 0 students

A sample "p" response (from student 4816):

Physics midterm question: tension forces on static beams

Physics 205A Midterm 2, Fall Semester 2009
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e, Problem 8.32

[10 points.] A uniform beam attached to a pivot is supported by a horizontal cable attached to its end, or a 45° cable attached to its midpoint. Which cable (if any) has a greater tension? Explain your answer using Newton's laws and properties of torque.

Solution and grading rubric:

• p = 10/10:
  Correct. Shows quantitatively from Newton's first law that the magnitude of the cw torques from either cable are identical (as they support the ccw weight torques of identical beams), such that F_1*r_perp1 and F_2*r_perp2, but since r_perp1 is larger (sqrt(2)*L) vs. r_perp2 (L/2), then F_1 < F_2. (Alternately may set F_perp1*L and F_perp2*(L/2) equal to each other, with F_perp1 = F_1*sin(45 degrees) and F_perp2 = F_2, such that (0.707)*F_1*L = (0.5)*F_2*L, proving that F_1 must be smaller than F_2.
  
• r = 8/10:
  As (p), but argument indirectly, weakly, or only by definition supports the statement to be proven, or has minor inconsistencies or loopholes. States that cable 2 has more tension as it has a smaller lever arm, but does not explicitly show that r_perp2 is smaller than r_perp1 (i.e., no clear distinction between r and r_perp).
  
• t = 6/10:
  Nearly correct, but argument has conceptual errors, or is incomplete. Garbled attempt at implementing Newton's first law for rotations and the F_perp*r or F*r_perp torque definition. May indicate that distance from CG as important.
  
• v = 4/10:
  Limited relevant discussion of supporting evidence of at least some merit, but in an inconsistent or unclear manner. May indicate that the length of the cable is important.
  
• x = 2/10:
  Implementation/application of ideas, but credit given for effort rather than merit.
  
• y = 1/10:
  Irrelevant discussion/effectively blank.
  
• z = 0/10:
  Blank.
  

Grading distribution:
Sections 70854, 70855, 72177
p: 5 students
r: 20 students
t: 32 students
v: 3 students
x: 1 student
y: 0 students
z: 0 students

A sample of a "p" response (from student 3026):