Question

A mass of 6 kg rests on a floor with a coefficient of static friction of 0.95 and coefficient kinetic friction of 0.30. One force is applied vertically downward of 22 Newtons and another force is applied at an angle of 28 degrees below the horizontal. Calculate what the magnitude this force needs to be in order to begin to move the mass. If that force is removed and another force is applied at an angle of 28 degrees above the horizontal (the downward force is still there), calculate the force needed to begin to move the force for that situation. How much more force is needed to begin the mass, in Newtons, when the force is angled below the horizontal compared to above the horizontal?

Answer 1

Here, u = coefficient of static friction = 0.95

for the force is applied below horizontal

for moving , Normal force

N = 6*9.81 + 22 + F * sin(theta)

N = 6*9.81 + 22 + F * sin(28)

for moving

F * cos(28) = u * N = 0.95 * (6*9.81 + 22 + F * sin(28))

solving for F

F = 175.80 N

the magnitude of force needed is 175.80 N

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when force is applied above horizontal

for moving , Normal force

N = 6*9.81 + 22 - F * sin(theta)

N = 6*9.81 + 22 - F * sin(28)

for moving

F * cos(28) = u * N = 0.95* (6*9.81 + 22 - F * sin(28))

solving for F

F = 57.80 N

the magnitude of force needed is 57.80 N

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more force needed = 175.80 N - 57.80 N = 118 N