Steel is very stiff, and Young’s modulus for steel is unusually large, $2\times {10}^{11}N/m$. A cube of steel 28 cm on a side supports a load of 85 kg that has the same horizontal cross section as the steel cube. (a) What is the magnitude of the normal force that the steel cube exerts on the load? (b) What is the compression of the steel cube? That is, what is the small change in height of the steel cube due to the load it supports? Give your answer as a positive number. The compression of a wide, stiff support can be extremely small.

The given data can be listed below as-

• The value of the Young’s modulus is$Y=2\times {10}^{11}N/m^2$ .
• The length of the steel is$L=0.28m$ .
• The mass of the object is $m=85kg$.

The Young’s modulus is described as the ratio of the stress of an object to the strain of that object.

The equation of the Young’s modulus gives the magnitude of the normal force and the compression of the steel cube.

(a)

The equation of the magnitude of the normal force can be expressed as-

$F_N=mg$

Here, $F_N$ is the normal force and m and g are the mass and the acceleration due to gravity.

Substituting the values in the above equation,

$$\begin{gathered} F_N=85kg\times 9.81m/s^2 \\ =833.85N \end{gathered}$$

Thus, the magnitude of the normal force is $833.85N$.

b)

the equation of the Young’s modulus of a wire can be expressed as-

$Y=\frac{FlA}{\left(∆LlL\right)}$

Here, Y is the Young’s modulus of a wire, F is the normal force, A is the area and$\left(∆L|L\right)$ is the change in the length.

The above equation can also be written as,

$$\begin{gathered} Y=\frac{\left(F|A\right)}{\left(∆L|L\right)} \\ =\frac{FL}{∆LA} \\ =\frac{FL}{∆L\left(L^2\right)}\left[A=L^2\right] \\ =\frac{F}{∆LL} \end{gathered}$$

Hence, further as,

$∆L=\frac{F}{YL}$

Substituting the values in the above equation,

$∆L=\frac{833.85N}{\left(2\times {{10}^{11}}^{N/m^2}\right)\left(0.28m\right)}$

Thus, the compression of the steel cube is $1.48\times {10}^{-8}m$.