A block a force 1705 N is shown on an incline of 29". A force 45 N is applied to it horizontally. 1705N Us=0.35 Uk =0.25
Posted: Wed May 04, 2022 9:17 am
A block a force 1705 N is shown on an incline of 29". A force 45 N is applied to it horizontally. 1705N Us=0.35 Uk =0.25 45N 29° a) Determine if the block is in equilibrium. The block is not in equilibrium v b) Find the magnitude and direction of the frictional force on the block. The magnitude of the force is The normal force is N and it is normal to the ramp c) If the block is in motion, determine whether it's moving up or down the ramp. The maximum friction force is Fmax = moving down the ramp N and it is up the ramp