Please show all the steps. Thank you. A force F is applied to a crate on an incl

Question

Please show all the steps. Thank you.

A force F is applied to a crate on an incline (see figure) so that the crate does not slide down the incline. The coefficient of static friction between the crate and the incline is 0.337. In the figure, = 33.9 . . A good free-body diagram will probably help you solve this problem. 3.37 kg What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline? (Turn in part (a) online.) N(±0.2N) Source: Serway and Faughn, College Physics, 5th edition, Problem 433

Explanation / Answer

component perpendicular to the plane:

mg cos(a) + P = R ...eq1

component parallel to the plane:
mg sin(a) = F ……..eq2

and

F = uR…………….eq3

by solving
(mg cos + F) = mg sin

F = (mg sin/ ) - mg cos

F = mg (sin/ – cos)

F = 3.37*9.81* (sin33.9/ 0.337 – cos 33.9)

F = 27.27483 N

F = 27.3 N

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