Question

A copper metal cube has each side of length $ \text{1m} $ . The bottom edge of the cube is fixed and tangential force $ \text{4}\text{.2 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{8}}}\text{N} $ is applied to a top surface. Calculate the lateral displacement of the top surface if modulus of rigidity of copper is $ \text{14 }\!\!\times\!\!\text{ 1}{{\text{0}}^{10}}\text{N/}{{\text{m}}^{2}} $ .

Answer 1

Hint: Modulus of rigidity is also known as the shear modulus of the body. It is given as the ratio between the shear stress to shear strain on a body. In this case, the students can directly use the formula for modulus of rigidity so as to find the tangential force acting on the cube.

Complete step-by-step answer:
We are given a copper metal cube whose side is given as 1m. We need to calculate the lateral displacement when a tangential force is applied on the top surface of the cube while its base is kept fixed.
Let us have a look at the formula for the modulus of rigidity or the shear modulus of a body.
If, $ \text{ }\!\!\eta\!\!\text{ } $ is the shear modulus of the body, $ \text{ }\!\!\lambda\!\!\text{ } $ is the lateral displacement, then we can write the formula as,
 $ \eta =\dfrac{F}{A\lambda } $
where, $ \text{F} $ is the force applied on the body and $ \text{A} $ is the area of the body.
Now, let us substitute the values given in the question in the above equation to obtain the value of the lateral displacement.
As the given body is a cube, so its area will be the cube of the length of its sides.
 $ \begin{align}
  & 14\times {{10}^{10}}=\dfrac{4.2\times {{10}^{8}}}{{{(1)}^{3}}\lambda } \\
 & \Rightarrow \lambda =0.003m \\
\end{align} $
Therefore from the above calculations, we can conclude that in case of the given question, the lateral displacement of the body is $ 0.003\text{m} $ .

Note: The students must be careful with the units of the parameters used in the formula. In order to correctly define a parameter, the unit is a very important element, and as such proper units should be used every time while solving a problem.