find the horizontal force necessary to hold the 68.9 kg crate at rest on the inclined plane of angle 20 degrees

Question

there is a crate on the inclined plane where the angle is 20 degrees . in order to hold the crate to rest what horizontal force is needed

 

Answer 1

The gravitational force on the crate is Fg = mg = (68.9)*(-9.8) = -675.22 newtons.  Click here for a diagram:

 

http://www.wyzant.com/resources/files/299638/crate_on_an_inclined_plane

 

The component of the force along the incline is mg*sin(20^o).  A horizontal force, F_H, must exactly counter the force pushing the crate down the incline to keep the crate at rest. So:

 

F_H*cos(20^o) = -mg*sin(20^o)

 

F_H = -mg*tan(20^o)

 

= -(-675.22)*(0.364)

 

= 245.76 newtons