A block weighing 70.0 N rests on a plane inclined at 25.0 Degree to the horizont

Question

A block weighing 70.0 N rests on a plane inclined at 25.0 Degree to the horizontal. A force F is applied to the object at 45.0 Degree to the horizontal, pushing it upward on the plane. The coefficients of static and kinetic friction between the block and the plane are, respectively, 0.378 and 0.156. (a) What is the minimum value of F that will prevent the block from slipping down the plane? 290 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. 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

a) the forces on the block are the component of gravity down the plane, which is mg sin(theta), and the component of friction up the plane, which is u mg cos(theta) (u=coeff of friction) and the force F up the plane

for the block to remain stationary, these forces must sum to zero, or:

F - mg sin(theta) + u mg cos(theta) = 0 > F=mg(sin(theta)-u cos(theta))

substitute values for m, g, theta and use ustatic since the block is not moving

b) F must exceed the value found in a)

c) F must exceed mg(sin(theta) - u cos(theta)) where u is now kinetic friction

substitute and check it out ur points deserve this only sorry for the inconvenience any doubts ping me

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