Question

Two bodies of masses m₁ and m₂ are connected a light string which passes over a frictionless massless pulley. If the pulley is moving upward with uniform acceleration `"g"/2`, then tension in the string will be ______.

Options

• `(3"m"_1"m"_2)/("m"_1+"m"_2)"g"`

• `("m"_1+"m"_2)/(4"m"_1"m"_2)"g"`

• `(2"m"_1"m"_2)/("m"_1+"m"_2)"g"`

• `("m"_1"m"_2)/("m"_1+"m"_2)"g"`

Answer 1

Two bodies of masses m₁ and m₂ are connected a light string which passes over a frictionless massless pulley. If the pulley is moving upward with uniform acceleration `"g"/2`, then tension in the string will be `underlinebb((3"m"_1"m"_2)/("m"_1+"m"_2))`.

Explanation:

The pulley is accelerating up with acceleration `"g"/2`.

We can consider masses m₁ and m₂ to be moving in accelerated frame of reference.

So a pseudo acceleration `"g"/2` acts on both m₁ and m₂ in downward direction.

Let m₁ moves up and m₂ moves down as seen from the pulley, and let the tension in string be T.

For m₁:

T - m₁ `("g"+"g"/2)`- m₁a                        ...(i)

similarly for mass m₂:

m₂g + m₂`"g"/2` - T - m₂a

⇒ m₂ `((3"g")/2)` - T - m₂a                      ...(ii)

using equations (i) and (ii)

(m₂ - m₁) `(3"g")/2-("m"_1+"m"_2)"a"`

⇒ a =  `(3("m"_2-"m"_1))/(2("m"_1+"m"_2))"g"`

Substituting for a in equation (i) and solving

T = `(3"m"_1("m"_2-"m"_1))/(2("m"_1+"m"_2))"g"+(3"m"_1"g")/2`

= `(3"m"_1"m"_2)/("m"_1+"m"_2)"g"`