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).
Step 1: Understand pulley system
The small pulley is attached to the block → it is a movable pulley
So, the block is pulled by two tensions:
Total pull on block = $2T$
Step 2: Forces on block (along incline)
Weight component down the plane:
$ 50g \sin 30^\circ = 50g \times \frac{1}{2} = 25g $
Normal reaction:
$ N = 50g \cos 30^\circ = 50g \times \frac{\sqrt{3}}{2} $
Friction (opposing upward motion):
$ f = \mu N = 0.3 \times 50g \times \frac{\sqrt{3}}{2} $
Step 3: Condition for upward motion
$ 2T = 25g + 0.3 \times 50g \times \frac{\sqrt{3}}{2} $
Simplify:
$ 2T = 25g + 15g \times \frac{\sqrt{3}}{2} $
Step 4: Relation with hanging mass
$ T = Mg $
So:
$ 2Mg = 25g + \frac{15\sqrt{3}}{2}g $
Cancel $g$:
$ 2M = 25 + \frac{15\sqrt{3}}{2} $
Step 5: Solve for $M$
$ M = \frac{1}{2}\left(25 + \frac{15\sqrt{3}}{2}\right) $
$ \sqrt{3} \approx 1.732 $
$ M = \frac{1}{2}(25 + 12.99) = \frac{37.99}{2} \approx 19.0 $
Final Answer:
$\boxed{19.0 \text{ kg}}$
Important Questions from Equilibrium and Friction
How does a lubricant reduce friction between moving parts of a machine?
The forces whose line of action lie along the same line are known as:
The necessary condition of equilibrium of a body is-
The forces which meet at one point and have their line of action in different planes are called
If in a planar system, only 2 reaction forces are acting, then the system is:-
