Forces on an Inclined Plane

Problem Overview

A crate of mass m = 100 kg is pushed at constant speed up a frictionless ramp (θ = 30°) by a horizontal force F. What are the magnitudes of (a) F and (b) the force on the crate from the ramp?

Solution to Part (a): Calculating the Pushing Force

So For part (a):

We start with the information given in the question that the crate moves at a constant speed, implying an acceleration (a) of 0 m/s² because if there is no change in speed, there is no acceleration.

So what we do is first resolve the force F into its components: one along the incline that is F cos(θ) and one perpendicular to it F sin(θ). Next we do the familiar thing of resolving mg into its respective components along the plane and perpendicular to the plane. So this become mg cos (θ) and this is mg Sin (θ). So now we have all the forces acting on the block along the plane and perpendicular to it. 

As a next step, we will apply Newton's second law along the incline, keeping in mind this as the sign notation, what we get is:

F cos(θ) - mg sin(θ) = m X 0 or

F cos(30°) - 100 * 9.8 sin(30°) = 0.

Solving for F, we F ≈ 566 N.

Solution to Part (b): Determining the Force from the Ramp

For part (b):

We need to find the normal force Fn, which is the support force from the incline on the crate. Remember the normal force always acts perpendicular to the surface to which the object is in contact with.

Now, since there is no acceleration perpendicular to the incline, Newton's second law in the perpendicular direction gives us: 

Fn – F Sin (θ) – mg Cos (θ) = 0, so you can see I have taken Fn as positive as it is acting in the +ve Y direction and these two forces as negative as they are acting in the negative Y direction…then if we substitute the values, we get 

Fn - 566 sin(30°) - 100 X 9.8 cos(30°) = 0 or

Fn = 100 * 9.8 cos(30°) + 566 sin(30°). 

That gives us Fn = 1133 N.

So, the normal force Fn is 1133 N

Key Takeaways

The key takeaway from this problem is that we first need to resolve all forces in the XX and YY plane then use Newton’s 2nd law of motion to find net force along each plane and then equate it with the product of mass and acceleration in the respective plane, keeping the sign notation in mind.