Example 5.5 from College Physics, 5.3 Elasticity: Stress and Strain

Calculating Force Required to Deform: That Nail Does Not Bend Much Under a Load

Find the mass of the picture hanging from a steel nail as shown in Figure 5.17, given that the nail bends only $1.80 µm$ . (Assume the shear modulus is known to two significant figures.)

Diagram showing the side view a nail in a wall, deformed by the weight of a picture hanging from it. The weight w of the picture is downward. There is an equal force w upward on the nail from the wall. The nail is 1 point five zero millimeters thick. The length of the nail that is outside the wall is five point zero zero millimeters. The deformation delta x of the nail as a result of the picture is 1 point eight zero micrometers.

Figure 5.17 Side view of a nail with a picture hung from it. The nail flexes very slightly (shown much larger than actual) because of the shearing effect of the supported weight. Also shown is the upward force of the wall on the nail, illustrating that there are equal and opposite forces applied across opposite cross sections of the nail. See Example 5.5 for a calculation of the mass of the picture.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
The force $F$ on the nail (neglecting the nail’s own weight) is the weight of the picture $w$ . If we can find $w$ , then the mass of the picture is just $\frac{w}{g}$ . The equation $Δ x = \frac{1}{S} \frac{F}{A} L_{0}$ can be solved for $F$ .
Solution
Solving the equation $Δ x = \frac{1}{S} \frac{F}{A} L_{0}$ for $F$ , we see that all other quantities can be found:
$F = \frac{SA}{L_{0}} Δ x .$
5.41

S is found in Table 5.3 and is $S = 80 \times 10^{9} {N/m}^{2}$ . The radius $r$ is 0.750 mm (as seen in the figure), so the cross-sectional area is
$A = {πr}^{2} = 1 . 77 \times 10^{- 6} m^{2} .$
5.42

The value for $L_{0}$ is also shown in the figure. Thus,
$F = \frac{(80 \times 10^{9} {N/m}^{2}) (1 . 77 \times 10^{- 6} m^{2})}{(5 . 00 \times 10^{- 3} m)} (1 . 80 \times 10^{- 6} m) = 51 N.$
5.43

This 51 N force is the weight $w$ of the picture, so the picture’s mass is
$m = \frac{w}{g} = \frac{F}{g} = 5 .2 kg .$
5.44
Discussion
This is a fairly massive picture, and it is impressive that the nail flexes only $1.80 µm$ —an amount undetectable to the unaided eye.

0 of 3 steps open

How did it go?

Strategy

Solution

Discussion

Strategy

Solution

Discussion