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Problem:
A horizontal force of 2N is just sufficient to prevent a block of mass 1kg from sliding down a rough plane inclined at arcsin7/25 to the horizontal. Find the coefficient of friction between the block and the plane and the acceleration 1.92m/s^2 simply because the mass if 1kg. Using F= ma. But the answer in the back of the book says it's 1.97 instead of 1.92, is this a printing error or have I gone wrong somewhere :S?

Answer:
θ =arcsin(7/25) = 16.26o

So summing forces along the plane ....F*cos(16.26) + f - m*g*sin(16.26) = 0 where f is the frictional force = µ*N

Summing force perpendicular we get N - F*sin(16.2) - m*g*cos(16.26) = 0

So N = F*sin(16.26) + m*g*cos(16.26)

Putting this all together

F*cos(16.26) + µ*(F*sin(16.26) + m*g*cos(16.26)) - m*g*sin(16.26) = 0

2*cos(16.26) + µ*(2*sin(16.26) + 1*9.8*cos(16.26)) - 1*9.8*sin(16.26) = 0

µ*(9.968) = 0.82396

So µ = 0.82396/9.968 = 0.0827

Now without the force we have m*g*sin(16.26) - µ*m*g*cos(16.26) = m*a

So a = g*(sin(16.26) - 0.0827*cos(16.26)) = 1.97m/s^2

Labels: Acceleration, Block

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