Two identical blocks A and B are connected by a rigid rod. The blocks rest against vertical and horizontal planes, as shown in the figure. The coefficient of static friction at the vertical and horizontal planes is the same. If the sliding impends when \(\theta = 45^{\circ}\), the value of the coefficient of static friction is (rounded off to two decimal places).
Show Hint
Analyze block A (against the wall) and block B (on the floor) separately using their own normal and friction forces; the rigid rod is a two force member acting along its own axis.
Step 1: Identify the forces acting on each block.
Block A rests against the vertical wall, so the wall exerts a normal force \(N_A\) on it that is horizontal, and a friction force \(f_A\) that acts vertically, since friction always acts along the contact surface, which is vertical here.
Block B rests on the horizontal floor, so the floor exerts a normal force \(N_B\) on it that is vertical, and a friction force \(f_B\) that acts horizontally.
The rigid rod AB carries no load other than at its two pinned ends, so it acts as a two force member. This means the rod can only push or pull along its own axis, at angle \(\theta\) to the horizontal, and it applies equal and opposite forces to blocks A and B along this line.
Step 2: Write equilibrium for block A.
Gravity pulls the whole assembly down, so block A tends to slide down along the wall. Friction \(f_A\) opposes this, so it acts upward on A.
Let the rod push block A away from B with force \(F\) along the rod axis. Resolving along the rod direction, the horizontal component of this force is \(F\cos\theta\) and the vertical component is \(F\sin\theta\), directed up and away from the wall corner.
Horizontal equilibrium of A: \(N_A = F\cos\theta\).
Vertical equilibrium of A: \(f_A + F\sin\theta = W\), where \(W\) is the weight of each block (the blocks are identical).
Since sliding is on the verge of happening, friction reaches its limiting value: \(f_A = \mu N_A = \mu F\cos\theta\).
Substituting this into the vertical equation:
\[ \mu F \cos\theta + F\sin\theta = W \]
\[ F = \frac{W}{\mu\cos\theta+\sin\theta} \]
Step 3: Write equilibrium for block B.
As A slides down, B tends to slide away from the wall (outward along the floor), so friction \(f_B\) acts toward the wall on B.
By Newton's third law, the rod pushes B with the same force \(F\), directed back toward A.
Vertical equilibrium of B: \(N_B = W + F\sin\theta\).
Horizontal equilibrium of B: \(F\cos\theta = f_B\), and at the verge of sliding, \(f_B = \mu N_B\). So:
\[ \mu(W+F\sin\theta) = F\cos\theta \]
Step 4: Combine the two block equations.
Divide the block B relation through by \(F\):
\[ \mu\left(\frac{W}{F}+\sin\theta\right) = \cos\theta \]
From Step 2, \(\dfrac{W}{F} = \mu\cos\theta+\sin\theta\). Substituting:
\[ \mu(\mu\cos\theta+\sin\theta+\sin\theta) = \cos\theta \]
\[ \mu^2\cos\theta + 2\mu\sin\theta = \cos\theta \]
Dividing throughout by \(\cos\theta\):
\[ \mu^2 + 2\mu\tan\theta - 1 = 0 \]
Step 5: Solve for the coefficient of friction at \(\theta = 45^{\circ}\).
At \(\theta = 45^{\circ}\), \(\tan\theta = 1\), so the equation becomes:
\[ \mu^2 + 2\mu - 1 = 0 \]
Using the quadratic formula:
\[ \mu = \frac{-2\pm\sqrt{4+4}}{2} = -1\pm\sqrt{2} \]
The coefficient of friction must be positive, so
\[ \mu = \sqrt{2}-1 = 0.4142 \]
Final Answer:
Rounded off to two decimal places, the coefficient of static friction is \(\mu \approx 0.41\).
\[ \boxed{\mu \approx 0.41} \]
