Question

If the tension in the string in the following figure is 16 N and the acceleration of each block is 0.5 m/s², find the friction coefficients at the two contact with the blocks.

Answer 1

From the free body diagram:
μ₁R + m₁a − F = 0
μ₁R + 1 − 16 = 0 (R = mg cos θ)

⇒ μ₁(2g) + (−15) = 0
`mu_1=15/20=0.75`
Again, 
μ₂R₁ + ma = F − mg sin θ = 0
μ₂R₁ + 4 × 0.5 = 16 − 4g sin 30° = 0
R₁ = mg cos θ      (θ = 30°)
⇒ μ₂ (20√3) + 2 + 16 - 20 = 0
`=> mu_2=2/20sqrt3=1/17.32`
= 0.057 = 0.06
Therefore, the friction coefficients at the two contacts with blocks are μ₁ = 0.75 and μ₂ = 0.06.