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

Question

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 35.0° to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.336 and 0.156. (a) What is the minimum value of F that will prevent the block from slipping down the plane? Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 100%. 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

Explanation / Answer

given:
weight = 87.5 N
angle = 25 degrees
Gravitational pull acting down = mg sin
Normal force = mg cos - F sin
Frictional force = (mg cos + F sin ) acts up when the block tends to come down and acts down when the block goes up.
Force acting up along the plane = F cos

(a) What is the minimum value of F that will prevent the block from slipping down the plane?
F cos + (mg cos + F sin ) = mg sin
F (cos + sin ) = mg (sin - cos )
F (cos 25 + 0.336 sin 25) = 87.5 (sin 35- 0.336*cos 35)
F = 24.9N

(b) What is the minimum value of F that will start the block moving up the plane?
F cos - (mg cos + F sin ) = mg sin
F (cos - sin ) = mg (sin + cos )
F (cos 25 -0.336 sin 25) = 87.5 *(sin 35+0.336*cos 35)
F = 97.17 N

(c) What value of F will move the block up the plane with constant velocity?
F cos - (mg cos + F sin ) = mg sin where = 0.156
F (cos 25 -0.156*sin 25) = 87.5 *(sin 35+0.156*cos 35)
F =73.N

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