A 4000N block is moved up a slope that is 20 degrees to the horizontal. If the coefficient of friction is 0.3, find the force parallel to the slope to: a) move the load up the slope b) prevent the load from sliding back

Answer 1

W=mg= 4000 or do i multiply this by 9.81

Answer 2

mg=4000 N is correct.
θ=20°
μ=0.3 (assumed static friction)

Gravity force along plane
F=mg sin(θ)

Friction force along inclined plane
=μmg cos(θ)
direction acts against movement.

To move load up slope:
Force required
=mg sin(θ)+μmg cos(θ)

To move load down slope:
Force required
=μmg cos(θ)-mg sin(θ)
if μmg cos(θ) > mg sin(θ)
=0 otherwise

Answer 3

F=4000sin(20)= 1368.1
FF= 0.3*4000cos(20)= 1127.6
FR= 4000sin(20) + 0.3*4000cos(20)= 2495.7
So 2495.7 N required to move it up the slope
1127.6 N is required to prevent the load from sliding back

Answer 4

Downward force due to gravity is
F=1368.1

Friction force resisting movement is
FF=1127.6

Therefore net downward force
=force required to prevent load from sliding down
=F-FF
=1368.1-1127.6
=240.45 N