Question

Two masses m₁ and m₂ are connected with a string passing over a frictionless pulley fixed at the comer of the table as shown in the figure. The coefficient of static friction of mass m₁ with the table is µ_s Calculate the minimum mass m₃ that may be placed on m₁ to prevent it from sliding. Check if m₁ = 15 kg, m₂ = 10 kg, m₃ = 25 and µ_s = 0.2.

 

Answer 1

Let m₃ is the mass added on m₁

Maximal static friction

`f_s^max` = µ_sN = µ_s (m₁ + m₃)g

Here,

N = (m₁ + m₃)g

Tension acting on string = T = m₂ g

Equate (1) and (2)

µ_s(m₁ + m₃) = m₂g

µ_sm₁ + µ_sm₃ = m₂

m₃ = `f_s^max`  – m₁

(ii) Given,

m₁ = 15 kg, m₂ = 10 kg : m₃ = 25 kg and µ_s = 0.2

m₃ = `f_s^max`  – m₁

m₃ = `10/0.2` – 15 = 50 – 15 = 35 kg

The minimum mass m₃ = 35 kg has to be placed on ml to prevent it from sliding. But here m₃ = 25 kg only.

The combined masses (m₁ + m₃) will slide.