A block of 50 kg mass is on an inclined plane. The block is connected to another hanging mass M by an inextensible massless string through two massless pulleys as shown in the figure below. The coefficient of static friction between the block and the inclined plane is 0.3. Neglecting pulley friction, the minimum value of M required to start the upward motion of the block is ________ kg (rounded off to 1 decimal place).
Show Hint
The movable pulley on the block gives it twice the string tension, so balance forces along the incline at the point of impending upward slip.
Step 1: Read the pulley arrangement.
The block carries a small movable pulley bolted to it, and the string runs from a fixed point, around this movable pulley, over the large fixed pulley at the top, and down to the hanging mass \(M\). Because the block's pulley is a movable pulley, two string segments pull the block toward the top of the incline, so the net pulling force on the block is \(2T\), where \(T\) is the tension in the string.
Step 2: Relate T to the hanging mass.
The string is massless and the pulleys are frictionless, so the tension is the same throughout the string. At the point of impending motion the hanging mass is in equilibrium, giving \(T = Mg\).
Step 3: Write the force balance on the block along the incline.
At the verge of sliding up, friction acts down the incline at its full static value. Along the incline: \(2T = mg\sin\theta + \mu mg\cos\theta\), where \(m = 50\) kg, \(\theta = 30^{\circ}\), and \(\mu = 0.3\).
Step 4: Substitute T = Mg and solve for M.
\[ 2Mg = mg(\sin\theta + \mu\cos\theta) \implies M = \frac{m(\sin\theta+\mu\cos\theta)}{2} \]
With \(\sin30^{\circ}=0.5\) and \(\cos30^{\circ}=0.866\): \(M = \dfrac{50(0.5 + 0.3\times0.866)}{2} = \dfrac{50(0.5+0.2598)}{2} = \dfrac{50\times0.7598}{2} = \dfrac{37.99}{2} = 19.0\) kg.
Final Answer:
The movable pulley doubles the pulling force for a given string tension, so the required hanging mass works out to about 19.0 kg.
\[ \boxed{M \approx 19.0\ \text{kg}} \]
