# The coefficients of friction between the block and the rail are μN = 0.30 and μk = 0.25. Knowing that θ =65° determine the smallest value of P required (a) to start the block up the rail, (b) to keep it from moving down.

Question-AnswerCategory: Engineering MechanicsThe coefficients of friction between the block and the rail are μN = 0.30 and μk = 0.25. Knowing that θ =65° determine the smallest value of P required (a) to start the block up the rail, (b) to keep it from moving down.
S cornwell asked 1 year ago

The coefficients of friction between the block and the rail are μN = 0.30 and μk = 0.25. Knowing that θ =65° determine the smallest value of P required (a) to start the block up the rail, (b) to keep it from moving down.

Step: 1

(a)
Draw the free body diagram of the block.

Step: 2

Assume that the force in the positive x-direction is positive.

Apply the equilibrium condition,

…… (1)
Assume that the force in the positive y-direction is positive.

Apply the equilibrium condition,

…… (2)

Step: 3

The maximum value of frictional force,

As the block is starting to move up the rail (impending motion), frictional force:

…… (3)
Substitute the value of N in equation (1) and solve.

…… (4)
Substitute the value of P and N in equation (2) and solve.

Substitute the value of F in equation (4) and solve.

Therefore, the smallest value of the force P required to start the block moving up the rail is .

Step: 4

(b)

Draw the free body diagram of the block when it will try to slip down.

Step: 5

Take the condition of equilibrium.
Assume that the force in the positive x-direction is positive.

Apply the equilibrium condition.
…… (5)

Step: 6

Assume that the force in the positive y-direction is positive.

Apply the equilibrium condition,

…… (6)
The maximum value of frictional force,

As the block needs to be kept from moving down, frictional force,

…… (7)

Step: 7

Substitute the value of N in equation (5) and solve.

…… (8)
Substitute the value of P and N in equation (6) and solve.

Substitute the value of F in equation (8) and solve.

Therefore, the smallest value of force P required to keep the block from moving down is .