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A 15-kg block is on a ramp which is inclined at 20° above the horizontal. It is


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A 15-kg block is on a ramp which is inclined at 20° above the horizontal. It is connected by a string to a 19-kg mass which hangs over the top edge of the ramp. Assuming that frictional forces may be neglected, what is the downward acceleration of the 19-kg block?



Nov 16th, 2017

Given m1 = 15 kg, m2 = 19 kg, θ = 20°. Find the acceleration a.

The system moves with a constant acceleration a, that is the net force acting on the first mass is

F1 = m1a and the net force acting on the second mass if F2 = m2a. The forces acting on the first mass are the weight W1 = m1g, the reaction N, and the tension force T1. Note that the force N is compensated by the normal component of the weight W1,n = m1gcos θ and the net force is directed along the inclined plane and equals F1 = m1a = T1 – m1gsinθ, where the latter is the second component of the weight W1.

For the second mass we have F2 = m2a = m2g– T2. Note that the string has no mass and therefore, the forces T1 and T2 have the same magnitude. By adding the two above equations, we get

m1a + m2a = m2g – m1gsinθ, and

a = (m2 – m1sinθ)g /(m1 + m2) = (19 – 15 sin20°) 9.81 / (19 + 15) = 4.0 m/s2 .


Apr 23rd, 2015

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