91Ó°ÊÓ

First, we will draw a free body diagram for both blocks A and B. The forces acting on block A are its weight \(W_A\) acting vertically downward, the friction force between blocks \(A\) and \(B\) denoted as \(f\), and the normal force \(N\) exerted by the wall. The forces acting on block B are its weight \(W_B\) acting vertically downward, the friction force between blocks \(A\) and \(B\) (opposite to the friction on A), and the normal force from the wall acting horizontally on block B.

Now we write the equilibrium equations for both blocks in the horizontal and vertical directions. For block A, Horizontal Equilibrium: \(N = f \sin\theta\) Vertical Equilibrium: \(W_A = f \cos\theta\) For block B, Horizontal Equilibrium: \(N = W_B\sin\theta\) Vertical Equilibrium: \(f = W_B\cos\theta\)

Now, we solve the equations for block B's horizontal equilibrium and vertical equilibrium to find the friction force direction. From block B's horizontal equilibrium, we have: \(N = W_B\sin\theta\) From block B's vertical equilibrium, we have: \(f = W_B\cos\theta\) Now, dividing the horizontal equilibrium equation by the vertical equilibrium equation, we get: \(\frac{N}{f} = \frac{W_B\sin\theta}{W_B\cos\theta}\) \(\frac{N}{f} = \tan\theta\) Now, if \(\tan\theta > 0\), then the friction force on block B due to block A must act upward because \(N > 0\). If \(\tan\theta < 0\), the friction force must act downward because \(N < 0\). If \(\tan\theta = 0\), then the friction force direction is not determined since \(N = 0\). Since the given problem states that the wall is smooth and the blocks are on an inclined plane, it is implied that \(\theta > 0\) or \(\theta < 180^\circ\). This implies that \(\tan\theta > 0\). Therefore, the friction force on block B due to block A must be acting in the upward direction. Hence, the correct option is: (A) Upward