Question

A light string going over a clamped pulley of mass m
supports a block of mass M as showing in the figure.
The force on the pulley by the clamp is given by
m
М
|
(A) V2 Mg (B) V2 mg
(C) g16M + m) + m² (D) 8V(M + m) + M?​

Answer 1

Hi there,

The figure is missing in your question. I have attached a figure (first figure) which satisfies the question and have solved it accordingly.

Answer: g√[{M+m}² + {M}²]

Explanation:

Referring to the 2nd figure attached, we can see that the distribution of forces i.e.,  

• The string attached to the pulley exerts a tension “T”

• Since the mass of the pulley is given as “m” & the mass of the block is given as “M”, so there will be a force due to weight in the downward direction i.e., “mg” & “Mg” respectively.

Now, referring to the 3rd figure attached, we can consider that at equilibrium

• T = Mg  …… (i)

• The resultant force exerted on the pulley by the clamp can be seen in the direction of “F”.

• The total force acting in the downward direction will be denoted as “F1” which is given as  F1 = (M+m)g ….. (ii)

Therefore,  

The resultant force “F” on the pulley by the clamp is,

= √$\sqrt{F1^2 + T^2}$

= √[{(M+m)g}² + {Mg}²]  ...... [substituting from (i) & (ii)]

= g√[{M+m}² + {M}²]

Hope this is helpful!!!