Block A (mass 25 kg) rests on block B (mass 40 kg), which rests on a horizontal floor. A string attached to A runs up to the left to a fixed vertical wall, making 20° with the horizontal. A horizontal force P applied to B pulls it to the right. The c…

Answer: \(P=241.4\) N (string tension \(T=59.8\) N).

\(W_A=25\times9.81=245.25\) N, \(W_B=40\times9.81=392.4\) N. When B is about to move right, A is held by the string, so friction is limiting at both surfaces.

FBD of A: \(W_A\) down, \(N_1\) up (from B), friction \(F_1=0.25N_1\) to the right (B tends to drag A to the right), string tension T up-left at 20°.

\(\sum F_y=0:\ N_1+T\sin20^\circ=245.25\) ...(1)

\(\sum F_x=0:\ 0.25N_1=T\cos20^\circ\Rightarrow T=0.2660N_1\) ...(2)

From (1): \(N_1+0.2660N_1\times0.3420=245.25\Rightarrow N_1(1.0910)=245.25\Rightarrow N_1=224.8\) N.

\(F_1=0.25\times224.8=56.2\) N, \(T=56.2/\cos20^\circ=59.8\) N.

FBD of B: \(W_B\) down, \(N_1\) down (from A), \(N_2\) up (floor), \(F_1\) to the left (from A), \(F_2=0.3N_2\) to the left (floor), P to the right.

\(\sum F_y=0:\ N_2=N_1+W_B=224.8+392.4=617.2\) N, so \(F_2=0.3\times617.2=185.2\) N.

\(\sum F_x=0:\ P=F_1+F_2=56.2+185.2=241.4\) N.