Question

A mass of 9 kg rests on a floor with a coefficient of static friction of 0.92 and coefficient kinetic friction of 0.37. One force is applied vertically downward of 22 Newtons and another force is applied at an angle of 26 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 26 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,

mass ,m = 9 kg

coefficient of static friction , us = 0.92

coefficient of kinetic friction , uk = 0.37

theta = 26 degree

downward force ,F1 = 22 N

when the force is applied below the horizontal

let the seccond force be F2

the normal force , N = (m * g + F1 + F2 * sin(theta))

for the block to move

F2 * cos(theta ) = us * N

F2 * cos(theta) = us * (m * g + F1 + F2 * sin(theta))

F2 * cos(26) = 0.92 * ( 9 * 9.81 + 22 + F2 * sin(26))

solving for F2

F2 = 204.78 N

the force applied below the horizontal is 204.78 N

when the force is applied above the horizontal

let the seccond force be F2'

the normal force , N = (m * g + F1 - F2' * sin(theta))

for the block to move

F2' * cos(theta ) = us * N

F2' * cos(theta) = us * (m * g + F1 - F2' * sin(theta))

F2' * cos(26) = 0.92 * ( 9 * 9.81 + 22 - F2' * sin(26))

solving for F2'

F2' = 77.9 N

the force applied above the horizontal is 77.9 N

the more force required , F = F2 - F2' = 126.85 N