Question

In the arrangement shown, the pulleys are fixed and ideal, the strings
are light. m1>m2 (m, and S is a spring balance which is itself massless.
The reading of S (in unit of mass) is :
​

Answer 1

Answer:

$\frac{2m_1m_2}{m_1+m_2}$

Explanation:

The tension in the strings will be the same

Let the tension is T

If Block of mass m1 moves downward with acceleration a and block of mass m2 moves upward with acceleration a then

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

$T-m_2g=m_2a$   .................(2)

Adding the above two equations

$m_1g-m_2g=m_1a+m_2a$

$a=\frac{m_1-m_2}{m_1+m_2}$

Putting the value of a in eq (1)

$m_1g-T=m_1\frac{m_1-m_2}{m_1+m_2}$

$T=m_1g-m_1\frac{m_1-m_2}{m_1+m_2}$

$\implies T=\frac{m_1^2+m_1m_2-m_1^2+m_1m_2}{m_1+m_2}g$

$\implies T=(\frac{2m_1m_2}{m_1+m_2})g$

Therefore, the reading in the spring balance will be

$\frac{2m_1m_2}{m_1+m_2}$

Hope this is helpful.

Answer 2

The acceleration of the system =( mm +m

2

​

m

1

​

+m

2

​

​

)g and hence the tension T in the string is (

m

1

​

+m

2

​

2m

1

​

m

2

​

​

)g. The reading of the spring balance is T (in units of force) and

g

T

​

(in units of mass).