Question

Three blocks of masses m₁ = 2 kg, m₂ = 3 kg and m₃ = 5 kg are suspended with string, which is passed over a pulley as shown in the figure. Calculate the tension T₁, T₂ and T₃, when the whole system (a) is moving upward with an acceleration of 2 ms^-2, and (b) is held stationary.

Answer 1

Here, m₁ = 2 kg, m₂ = 3 kg and m₃ = 5 kg

(a) When the system is moving upward with an acceleration 'a' where a = 2 ms^-2.

As per the fig., consider the free body diagram of the system of three blocks. Then,

T₁ - (m₁ + m₂ + m₂) g = (m₁ + m₂ + m₃) a

or, T₁ = (m₁ + m₂ + m₃) (a + g)    ...(i)

or, T₁ = (2 + 3 + 5) x (2 x 9.8) = 118 N

Consider the free body diagram of blocks of masses m₂ and m₃. Then

T₂ - (m₂ + m₃)g = (m₂ + m₃) x a     ....(ii)

or, T₂ = (m₂ + m₃)(a + g)

⇒ T₂ = (3 + 5) x (2 x 9.8) = 94.4 N

Consider the free body diagram of block of mass m₃. Then,

T₃ - m₃g = m₃a    ....(iii)

or, T₃ = m₃(g + a) = 5 x (2 + 9.8) = 59 N

(b) When the system is held stationary:

Obviously, a = 0, setting a = 0, in the equations (i), (ii) and (iii), we have

T₁ = (m₁ + m₂ + m₃) g = (2 + 3 + 5) x 9.8

= 98 N

T₂ = (m₂ + m₃) g = (3 + 5) x 9.8 = 78.4 N

and T₃ = m₃g = 5 x 9.8 = 49 N