Answer to Question #110360 in Physics for Jess

Question

A 18.5 cm long probe used in a diagnostic process consists of a thin 16.0 cm long plastic rod with a cross-sectional area of and a 2.50 cm long nylon tip. Both the rod and tip have a cross-sectional area of 0.955×10−6 m2.

The (compressive) Young's modulus of plastic is 3.51 GPa, and the compressive Young's modulus of the nylon used is 2.02 GPa.

The instrument is pressed against a bone with a compressive force of 105 N. What is the change in length of the instrument? (to 2 s.f and in mm)

Answer 1

$$\Delta l_1=l_1\frac{F}{AY_1}$$

$$\Delta l_1=16\frac{105}{0.955\cdot 10^{-6}\cdot 3.51\cdot 10^9}=0.501\ cm$$

$$\Delta l_2=l_2\frac{F}{AY_2}$$

$$\Delta l_2=2.5\frac{105}{0.955\cdot 10^{-6}\cdot 2.02\cdot 10^9}=0.136\ cm$$

$$\Delta l=5.01+1.36=6.4\ mm$$