Question

In the following figure shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is μ₁, and that between the block of mass 4.0 kg and incline is μ₂. Calculate the acceleration of the 2.0 kg block if (a) μ₁ = 0.20 and μ₂ = 0.30, (b) μ₁ = 0.30 and μ₂ = 0.20. Take g = 10 m/s².

Answer 1

(a) From the free body diagram

R = 4g cos 30°
`=>"R"=4xx10xxsqrt3/2`
`= 20sqrt3"N"`                               (1)
μ₂R + m₁a − p − m₁g sin θ = 0
μ₂R + 4a − p − 4g sin 30° = 0              
⇒ 0.3 x (40) cos 30° + 4a − p − 40 sin 30° = 20             (2)

R₁ = 2g cos 30° = 10√3                       (3)
p + 2a − μ₁R₁ − 2g sin 30° = 0               (4)
From Equation (2),
`6sqrt3+4a-p-20=0`
From Equation (4),
`p+2a+2sqrt3-10=10`

`6sqrt3+6a+30+2sqrt3=0`

`=>6a=30-8sqrt3`

=30 - 13.85=16.15

`=>a=16.15/6`

= 2.69 = 2.7 m/s²

^{(b) In this case, the 4 kg block will move at a higher acceleration because the coefficient of friction is less than that of the 2 kg block. Therefore, the two blocks will move separately. By drawing the free body diagram of 2 kg mass, it can be shown that a = 2.4 m/s2.}