Question

Two cylinders are kept in a container as shown in the figure below. Top cylinder has a weight W and bottom cylinder has a weight of 2 W.



What is the reaction at A? (Neglect friction)

• A. 3 W
• B. 2.8 W
• C. 2.5 W
• D. $\frac{3}{4}$ W

Answer 1

The correct option is D $\frac{3}{4}$ W



$cos\alpha =30/50$



F.B.D. of top cylinder gives W = R sin $\alpha$

F.B.D. of bottom cylinder gives

$N_2=Rcos\alpha$

From the above relations, we get

$tan\alpha =\frac{W}{N_2}$

$tan\alpha =\frac{40}{30}$

$\frac{W}{N_2}=\frac{40}{30}$

$N_2=\frac{3}{4}W$