Question

in the figure shown the string and pulley are ideal and F acts vertically. The magnitude of F for the blocks to be in equilibrium is​

Answer 1

answer : option (c) $(m_2-m_1)g$

explanation : draw free diagram and then solve.

for block of mass $m_1$,

F + $m_1g$ - T = $m_1a$...(1)

for block of mass $m_2$,

T - $m_2g$ = $m_2a$....(2)

solving equations (1) and (2),

F + $m_1g-m_2g$ = $(m_1+m_2)a$......(3)

a/c to question, system is in equilibrium.

so, a = 0

then from equation (3),

$F+(m_1-m_2)g=0$

or, $F=(m_2-m_1)g$

Answer 2

Answer:

Explanation:

In any case of the pulley first we have to see the line of application of force which in here is on m1 hence we have to balance the other mass with respect to m2.

So, in order to balance the force we know that the magnitude of force will be greater than the amount of tension for m1. While in m2 the magnitude of tension will be greater than the amount of its own weight.

Hence, by that case taking the acceleration to be in downwards direction as the force applied is also in downwards direction.

Let the tension in both of the strings be T.

So, for m1 ... F + m1g = T.

And for m2...T - m2g = 0 , no acceleration to make it equilibrium.

By equating the equations with tension,

We get , F + m1g = m2g.

              F = g( m2 - m1).

Option C is the correct answer.