Question

A block weighing 87.5 N rests on a plane inclined at 25.0° to the horizontal. A...

A block weighing 87.5 N rests on a plane inclined at 25.0° to the horizontal. A force F is applied to the object at 50.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.377 and 0.156.

(a) What is the minimum value of F that will prevent the block from slipping down the plane?
N
(b) What is the minimum value of F that will start the block moving up the plane?
N
(c) What value of F will move the block up the plane with constant velocity?
N

Homework Answers

Answer #1

here,

weight , W = 87.5 N

theta = 25 degree

phi = 50 degree

a)

the minimum value of F that will prevent the block from slipping down the plane be F

F * cos(phi) + us * ( m * g * cos(theta) - F * sin(theta) ) = m * g * sin(theta)

F * cos(50) + 0.277 * ( 87.5 * cos(25) - F * sin(50)) = 87.5 * sin(25)

solving for F

F = 34.9 N

b)

the minimum value of F that will prevent the block from moving up the plane be F

F * cos(phi) = us * ( m * g * cos(theta) - F * sin(theta) ) + m * g * sin(theta)

F * cos(50) = 0.377 * ( 87.5 * cos(25) - F * sin(50)) + 87.5 * sin(25)

solving for F

F = 71.8 N

c)

let the minimum force be F

F * cos(phi) = uk * ( m * g * cos(theta) - F * sin(theta) ) + m * g * sin(theta)

F * cos(50) = 0.156 * ( 87.5 * cos(25) - F * sin(50)) + 87.5 * sin(25)

solving for F

F = 64.7 N

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