Answer to Question #158903 in Physics for Annie

Question

A cylindrical rod of length 2.0 m and cross-sectional area 8.0 × 10^–5 m2 is made of hard steel. The rod is placed under a tensile stress of 200 MPa.

Which one of the following is the best estimate of the force that must be applied to this hard steel rod to provide this stress?

Answer 1

By definition, the stress is:

$$\sigma = \dfrac{F}{S}$$

where $F$ is the force that must be applied to this hard steel rod, and $S = 8\times 10^{-5}m^2$ is its cross-sectional area. Since $\sigma = 200\times 10^6Pa$, the force is:

$$F = \sigma S\\ F = 200\times 10^6\cdot 8\times 10^{-5} = 16000N$$

Answer. 16000 N.