A 20.0kg block rests on a plane inclined at 20.0o to the horizontal. The coefficient of static friction is 0.250, and the coefficient of kinetic friction is 0.150. (a) What is the minimum magnitude of the force F, parallel to the plane that will prevent the block from slipping down the plane? (b) What value of F is required to move the block up the plane at constant velocity?
(a)To move the block, the force F must equal or exceed the sum of the graviational component down the ramp and the friction force
F = (mu,s)cos 20 * M g + Mg sin 20
= Mg (0.25*0.9397 + 0.3420)
Complete the calculation using M and g.
(b) The (lower) force F must equal the sum of kinetic friction and the weight component
F = Mg (0.15 cos 20 + sin 20)
i got 113.07 for part A.
but the answer should be 21.0 N.
but i don't understand how the book got that answer.
I was wrong; the book is correct. When static friction is resisting motion, the friction force is in the same direction as the applied force. Therefore
F = -(mu,s)cos 20 * M g + Mg sin 20
= Mg (-0.25*0.9397 + 0.3420) = 21.0 N
ohhh
thank you SO MUCH!
1 answer