Question

30. A wire ABC supports a body of mass m as shown in the figure. The wire passes over a fixed pulleys at B and is firmly attached to a vertical wall at A. The line AB makes an angle • with the vertical, and the pulley at B exerts on the wire a force F, then the value of force F (a) 2mgsin theta/2 b)2mg cost theta/2. c)3mgcos theta/2. d) none​

Answer 1

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Answer 2

Given:

A pulley-mass system as shown in figure

The angle between AB and vertical = ∅

To find:

Value of F

Solution:

Forces acting on point B are Tension (T), and mg (due to body)

Since point B is in equilibrium, the vector sum of all these forces should be equal to F

If F1 and F2 are horizontal and vertical components of F respectively, then

F1 = T sin∅

F2 =mg - T cos ∅

F = $\sqrt{F1^2 + F2^2}$

= $\sqrt{T^2sin^2\phi + m^2g^2 + T^2cos^2\phi-2 mg T cos\phi}$

Since sin²Ф+cos²Ф =1

F = $\sqrt{T^2 + m^2g^2-2mgTcos\phi}$

= 2mg cos ∅/2

Hence, the force F = 2mg cos ∅/2