Answer to Question #132411 in Mechanics | Relativity for K.Arun sriram

Question

A 50 Kg block is attached to mass M using string which is passing over a smooth pulley .Determine the range of mass M that will keep the block in equilibrium.Assume static friction is 0.03 and angle is 30

Answer 1

If block is moving down the inclined plane,

then equation of motion will be

$Ma = Mgsin\theta -f -T$ (1)

where $f$ is friction force.

$50a = T - 50g$ (2)

Since block is in equilibrium, then a = 0,

Hence solving these equations we get

$Mgsin\theta -\mu Mgcos\theta = 50 g \implies M = \frac{50}{sin\theta - \mu cos\theta} = 105.5 kg$

If block moves up along inclined plane,

$Ma =T- Mgsin\theta -f$ (3)

$50a = 50g - T$ (4)

Since a=0, then

$50g = Mgsin\theta + \mu Mgcos\theta \implies M = \frac{50}{sin\theta + \mu cos\theta} = 95.06 kg$

So range for M is $95.06kg \leq M \leq 105.5 kg$