Suppose that you are going to measure Young’s modulus for three rods by measuring their stretch when they are suspended vertically and weights are hung from them (Figure 4.56)

(Figure 4.56)

Rod 1 is 2.7 m long and cylindrical with radius 4 mm (1 mm is 0.001 m). Rod 2 is 3.2 m long by 12 mm wide by 6 mm deep. Rod 3 is 3 m long by 6 mm wide by 6 mm deep. The definition of Young’s modulus, Y = (F/A)/(∆L/L), includes the quantity A, the cross-sectional area.

(a) What is the cross-sectional area of rod 1?

(b) What is the cross-sectional area of rod 2?

(c) What is the cross-sectional area of rod 3?

Short Answer

Expert verified

The results are 5.03×105m2,7.2×105m2 and3.6×105m2 .

Step by step solution

01

Given data

Rod 1,

L = 2.7 m

R = 4mm = 0.004 m

Rod 2,

L = 3.2 m

w = 12mm = 0.012 m

d = 6mm = 0.006 m

Rod 3,

L = 3 m

w = 0.006 m

d = 0.006 m

02

(a)Determine the cross-sectional area of rod 1 

A1=πr2=3.14×(0.004)2=5.0265×105m2A15.03×105m2

03

(b) Determine the cross-sectional area of rod 2 

A2=dw=0.012×0.006=7.2×105m2

04

(c)Determine the cross-sectional area of rod 3

A3=wd=0.006×0.006=3.6×105m2

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In outer space, a rod is pushed to the right by a constant force F (Figure 4.54). Describe the pattern of interatomic distances along the rod. Include a specific comparison of the situation at locations A, B, and C. Explain briefly in terms of fundamental principles.

Figure 4.54

Hint: Consider the motion of an individual atom inside the rod, and various locations along the rod.

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