Steel is very stiff, and the Young's modulus for steel is unusually large, 2.0*1011 N/m2. A cube of steel 25 cm on a side supports a load of 85 kg that has the same horizontal cross section as the steel cube. What is the magnitude of the "normal" force that the steel cube exerts on the load? By how much does the steel cube compress, due to the load that it supports? (Give your answer as a positive number.)

Force (normal) = 833.85 N Compression = 1.67 x 10⁻⁸ m

Explanation:

Given data Young's Modulus (Y) = 2 x 10¹¹ N/m², Length of one side of cube = 25 cm = 0.25 m, mass of load = 85 kg

Normal force is the force exerted upon an object that is in contact with another stable object. This force would be applied by the surface onto the object in the same vector and is used to keep the object stable while it rests on a surface.

We know from Newton's Second Law that

F = ma where m is the mass and a is the acceleration (in this case due to gravity) hence, the normal opposing force to the load applied by the surface would be equal to the force applied on the surface by the weight of the load on the surface, So

F (normal) = M (load) x a = 85 x 9.81 = 833.85 N

Compression is the change in length of an object by the exertion of force upon it. Using the Young's Modulus formula we can find this change in the cube of steel. The Young's Modulus is given by

Y = (F/A) / (ΔL/L), where Y is the Young's Modulus, F is the Force being applied on the object, A is the cross sectional area on which the said force is applied, ΔL is the change in length due to said force being applied and L is the original Length of the side of the cross sectional area.

Solving this for ΔL, we can re - arrange the equation

ΔL = (F x L) / (Y x A) since area of square is L x L we can simplify the equation to get

ΔL = (F) / (Y x L), substitute the values

ΔL = (833.85) / (2 x 10¹¹ x 0.25) = 1.67 x 10⁻⁸m