Question

A highly rigid cubical block A of small mass M and side L is fixed rigidly on another cubical block B of same dimensions and of low modulus of rigidity η, such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A at the top. The side of B turns by an angle $\phi$. Now the whole arrangement is turned upside down so that lower face of A is rigidly held to the horizontal surface and we apply a small force F perpendicular to the side face of B at the top. The Side of B now turns by an angle.

• A. $\phi$
• B. $\phi/2$
• C. $\phi/3$
• D. Zero

Answer 1

Answer: (A)

$\phi$

Answer 2

In the first scenario, block B is at the bottom and block A is on top. A force F is applied to the top of block A, causing a shear stress on block B. The shear stress is $\tau_1 = F/L^2$. The shear strain is $\gamma_1 = \tau_1/\eta = F/(\eta L^2)$. For small angles, the angle of turn $\phi$ is approximately equal to the shear strain: $\phi \approx \gamma_1 = F/(\eta L^2)$.

In the second scenario, the arrangement is inverted. Block A is now at the bottom, fixed to the horizontal surface, and block B is on top. The force F is applied to the top of block B. This again causes a shear stress on block B. The shear stress is $\tau_2 = F/L^2$. The shear strain is $\gamma_2 = \tau_2/\eta = F/(\eta L^2)$. Let the new angle of turn be $\phi_2$. For small angles, $\phi_2 \approx \gamma_2 = F/(\eta L^2)$.

Comparing the two angles, we see that $\phi = F/(\eta L^2)$ and $\phi_2 = F/(\eta L^2)$. Therefore, $\phi_2 = \phi$. The angle of turn remains the same regardless of which block is on top, as long as the force is applied to the upper block and the lower block is fixed.