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The mass of block in the figure is 9 kg. The coeff
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Answer #1

Given Data

mass m = 9 kg

coefficient of static friction u = 0.5

Angle theta = 30 deg

down force (weight) = friction force + Fv

Where friction force = coefficient of friction times the normal force (or the normal to the wall component of the applied force F).

So we have
Fv = Fcos(30)


W = uFsin(30) + Fcos(30)

m*g = uFsin(30) + Fcos(30)

m*g = F*[usin(30) + cos(30)]

F = m*g / (usin(30) + cos(30))

F = 9*9.81 / [0.50*sin(30) + cos(30)]

F = 88.3 / [1.116]

F = 79.11 N

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