5.3 Elasticity: stress and strain  (Page 6/15)

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 [link] , given that the nail bends only $\text{1.80 µm}$ . (Assume the shear modulus is known to two significant figures.)

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 $\mathrm{\Delta }x=\frac{1}{S}\frac{F}{A}{L}_0$ can be solved for $F$ .

Solution

Solving the equation $\mathrm{\Delta }x=\frac{1}{S}\frac{F}{A}{L}_0$ for $F$ , we see that all other quantities can be found:

S is found in [link] and is $S=\text{80}\times {\text{10}}^9{\text{N/m}}^2$ . The radius $r$ is 0.750 mm (as seen in the figure), so the cross-sectional area is

The value for $L_0$ is also shown in the figure. Thus,

This 51 N force is the weight $w$ of the picture, so the picture’s mass is

Discussion

This is a fairly massive picture, and it is impressive that the nail flexes only $\text{1.80 µm}$ —an amount undetectable to the unaided eye.