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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.
Solution
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.
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