force of a box and a slippery ramp

Question

A 10 kg box is being pushed up a slippery ramp. The coefficient of friction between the box and the ramp is just µ=0.1 ... (ramp is 10°)

a) what force does the man need to apply to the box to keep it traveling up the ramp at a steady speed?

b) what fraction of the weight force of the box is this ?

If the angle of the ramp is 45° then:

c) What force does the man need to apply to the box to keep it traveling up the ramp at a steady speed now?

d) What fraction of the weight force of the box is this?

Answer 1

Set up your free-body diagram something like the above.

There are 4 forces, F_F = Force of friction, F_P = Force of pushing, F_N = The Normal force, and W = weight (m*g).

All the forces are lined up with the coordinate axis I established except W, so we must use trig to break W into its components that line up with the coordinate axis so W_y = mgcos(10°) and W_x = mgsin(10°)

∑F_y = 0 because there is no motion in that direction.

∑F_y = F_N - W_y

F_N = W_y = mgcos(10°) = 10*9.8*cos(10°) = 96.51

F_F = μF_N = 0.1*96.51 = 9.651

∑F_x = F_P - F_F - W_x so F_P = F_F + W_x = 9.651 + 10*9.8*sin(10°) = 26.671 so F_P = 27 N

The fraction of the weight would be 26.671/98 = 0.272 or 27/100

For 45°, you just repeat the above using the angle of 45° in place of 10°. I got an answer of 76.23 N for the F_P