Physics quiz question: concrete sample compression

Physics 205A Quiz 6, fall semester 2017
Cuesta College, San Luis Obispo, CA

A concrete sample was (non-destructively) tested by compressing it with a stress of 9.9×10⁶ N/m²[*]. The Young's modulus of this concrete is 2.8×10¹⁰ N/m²[**]. If the concrete sample started with a height of 0.305 m, during testing it was compressed by:
(A) 1.1×10^–11 m.
(B) 1.1×10^–4 m.
(C) 3.5×10^–4 m.
(D) 5.5×10^–3 m.

[*] youtu.be/iCWsDHhbi9g.
[**] engineeringtoolbox.com/concrete-properties-d_1223.html.

Correct answer (highlight to unhide): (B)

Hooke's law is given by:

(F/A) = Y·(∆L/L),

where the compressive stress (F/A) is given as 9.9×10⁶ N/m². The amount that sample would be compressed by will be:

∆L = (F/A)·(L/Y),

∆L = (9.9×10⁶ N/m²)·((0.305 m)/(2.8×10¹⁰ N/m²)),

∆L = 0.0001078392857 m,

or to two significant figures, the concrete sample would compress by 1.1×10^–4 m.

(Response (A) is L/Y; response (C) is the strain (∆L/L) = (F/A)/Y; response (D) is the volume of the concrete cylinder, which cannot be determined from the values given above.)

Sections 70854, 70855
Exam code: quiz06Ho0k
(A) : 4 students
(B) : 40 students
(C) : 1 student
(D) : 0 students

Success level: 89%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.27

Physics quiz question: femur compression

Physics 205A Quiz 6, fall semester 2015
Cuesta College, San Luis Obispo, CA

An average femur bone (Young's modulus 9.4×10⁹ N/m²) has a length of 0.48 m and an approximate cross-sectional area of 1.72×10^–3 m², and reportedly can support a maximum 2.4×10⁴ N of force.[*][**] When this force is applied, the femur would compress by:
(A) 2.1×10^–9 m.
(B) 9.1×10^–9 m.
(C) 7.2×10^–8 m.
(D) 7.1×10^–4 m.

[*] "30 times the weight of an adult," orthopaedicsone.com/display/Review/Femur.
[**] "Weight of average adult: 178 lbs," wolfr.am/8dc2Ohi7.

Correct answer (highlight to unhide): (D)

Hooke's law is given by:

(F/A) = Y·(∆L/L),

such that the amount that the femur would be compressed is:

∆L = (F·L)/(A·Y),

∆L = ((2.4×10⁴ N)·(0.48 m))/((1.72×10^–3 m²)·(9.4×10⁹ N/m²)),

∆L = 0.0007125185552 m,

or to two significant figures, the femur would compress by 7.1×10^–4 m.

(Response (A) is (F·L·A)/Y; response (B) is (F·A)/(L·Y); response (C) is A/F.)

Sections 70854, 70855, 73320
Exam code: quiz06m45S
(A) : 3 students
(B) : 0 students
(C) : 3 students
(D) : 67 students

Success level: 92%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.17

Physics quiz question: stretching fishing lines

Physics 205A Quiz 6, fall semester 2014
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e Conceptual Question 11.4, Problems 11.1, 11.3

"Untitled"
Eileen Delhi
flic.kr/p/fGR8Fi

Trilene® XL Super Strong fishing line (Young's modulus 2.0×10⁹ N/m²) and Eagle Claw® Sportfisher fishing line (Young's modulus 3.1×10⁹ N/m²) have the same 10.0 m length [*]. The Trilene® fishing line has a cross-sectional area 1.8 times that of the Eagle Claw®. Both fishing lines are stretched with a tension force of 98 N. The __________ fishing line will stretch more.
(A) Trilene®.
(B) Eagle Claw®.
(C) (There is a tie.)
(D) (Not enough information is given.)

[*] S. Ottolini, G. Halpin, P. LaBruzzo, "Tensile Strength of Fishing Line," santarosa.edu/~yataiiya/E45/PROJECTS/Tensile%20Strength%20of%20Fishing%20Line%20Power%20Point.ppt.

Correct answer (highlight to unhide): (B)

Hooke's law for the Trilene® and Eagle Claw® fishing lines are given by:

(F/A_Tri) = Y_Tri·(∆L_Tri/L),
(F/A_Eagle) = Y_Eagle·(∆L_Eagle/L),

where tension F and the original, unstretched length L are the same for both fishing lines. The Trilene® fishing line has a cross-sectional area 1.8× that of the Eagle Claw® fishing line:

A_Tri = 1.8·A_Eagle.

The amount that the Trilene® fishing line will be stretched is given by:

∆L_Tri = (F·L)/(A_Tri·Y_Tri),

∆L_Tri = ((98 N)·(10.0 m))/((1.8·A_Eagle)·(2.0×10⁹ N/m²)),

∆L_Tri = (2.7×10^–7 m³)/A_Eagle.

Similarly, the amount that the Eagle Claw® fishing line will be stretched is given by:

∆L_Eagle = (F·L)/(A_Eagle·Y_Eagle),

∆L_Eagle = ((98 N)·(10.0 m))/((A_Eagle)·(3.1×10⁹ N/m²)),

∆L_Eagle = (3.1×10^–7 m³)/A_Eagle.

Thus this sample of Eagle Claw® fishing line will stretch more than the Trilene® fishing line sample.

Sections 70854, 70855, 73320
Exam code: quiz06eAg7
(A) : 13 students
(B) : 49 students
(C) : 2 students
(D) : 0 students

Success level: 77%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46

Physics quiz question: Miley Cyrus' "Wrecking Ball"

Physics 205A Quiz 6, fall semester 2013
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 2/e Conceptual Question 10.4

Miley Cyrus, "Wrecking Ball"
Terry Richardson (director)
Vevo, September 9, 2013

In a recent music video[*], pop singer Miley Cyrus sits on a (fake) demolition wrecking ball, together approximated as a 120 kg point mass hanging from a steel cable. While stationary, the steel cable stretches by 1.2 mm when 120 kg is hanging from it. If 240 kg were suspended from a steel cable of the same length with twice the radius, then it would stretch by __________ 1.2 mm.
(A) less than.
(B) exactly.
(C) more than.
(D) (Not enough information is given.)

[*] Don't bother watching it. Also the chain in the video is simplified here as a uniform cable.

Correct answer (highlight to unhide): (A)

Hooke's law for the thin and thick cables are given by:

(F_thin/A_thin) = Y·(∆L_thin/L),
(F_thick/A_thick) = Y·(∆L_thick/L),

where the Young's modulus Y and the original, unstretched length L are the same for the thin and thick cables (being both made of steel). The thick cable has twice the load of the thin cable:

F_thick = 2·F_thin.

The radii and thus the cross-sectional areas of the thin and the thick cables are also different:

A_thin = π·r_thin²,
A_thick = π·r_thick².

The thick cable has a radius twice that of the thin cable (r_thick = 2·r_thin), which will give it a cross-sectional area of four times that of the thin cable:

A_thick = π·r_thick² = π·(2·r_thin)² = 4·π·r_thin² = 4·A_thin.

Then setting the ratio of Y/L for the thin and thick cables equal to each other:

Y/L = Y/L,

F_thin/(A_thin·∆L_thin) = F_thick/(A_thick·∆L_thick),

and substituting in F_thick = 2·F_thin and A_thick = 4·A_thin:

F_thin/(A_thin·∆L_thin) = 2·F_thin/(4·A_thin·∆L_thick),

1/∆L_thin = 1/(2·∆L_thick),

thus:

∆L_thick = (1/2)·∆L_thin,

such that the thick cable will stretch less than the thin cable.

Sections 70854, 70855, 73320
Exam code: quiz06wR3k
(A) : 12 students
(B) : 32 students
(C) : 20 students
(D) : 0 students

Success level: 19%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.08