The ramp (picture link) is 18 meters long and 4.5 meters high. What force, parallel to the ramp (Fa) is ...?

Question

The ramp (picture link) is 18 meters long and 4.5 meters high.

*picture link: http://img206.imageshack.us/img206/2450/scangd4.jpg

a.What force, parrallel to the ramp (Fa) is required to slide a 25 kg box at constant speed to the top of the ramp if friction is disregarded?
b. What is the ideal Mechanical Advantage of the ramp?
c. What are the real Mechanical Advantage and the efficiency of the ramp if a parrallel force of 75 N is actually required?

Please help me with this problem.
I'm having trouble understanding it.

I asked this question before, but I was unable to get any help.

Answer 1

a. If theta is the angle between the slope and horizontal, you will need sin(theta)
Sin(theta) = 4.5 m/18 m.
The weight of the box, 25*9.8 Newtons, can be resolved into 2 perpendicular forces:

The component of the weight pointed downslope:
Fds = 25*9.8 N*sin(theta)

The other perpendicular force is usually called the normal force, Fn. It is perpendicular to the surface of the slope and would be needed if we were supposed to consider friction. In this case, we don't need it.

According to Newton's laws, the upslope force, FA, required to maintain a speed upslope is a force equal and opposite to the downslope force. I called it Fds.
So FA = -Fds

b. Since we ignored friction, our results apply to the ideal case. The mechanical advantage is weight/FA.

c. In this case it would be weight/75 N.