Find the normal force on an inclined plane.

[dirgenowitzki]

Homework Statement

Find the normal force acting on a mass of 5kg on an incline plane of 30 degrees.

Homework Equations

Fn = mgsin30 or mgcos30 ?

The Attempt at a Solution

I'm not sure if I use sin or cos when finding the normal force on an incline. I have trouble visualizing the angle perpendicular to the plane and moving the angles around.

 

[Redbelly98]

Welcome to Physics Forums.

If the surface makes an angle θ to the horizontal, then the normal makes the same angle θ from the vertical. Does that help your visualizing?

If it's still unclear, think about a horizontal surface, so that θ is zero, and answer these questions:

1. What is the normal force when the surface is horizontal? Draw a force diagram for yourself, if you need to, to figure this out.

2. Does your answer to (1) agree with mgsin0 or mgcos0?

Hope that helps!

 

[dirgenowitzki]

It works with mgsin theta, but when i try to draw it out it doesn't make sense to me.

 

[Doc Al]

You may want to read this tutorial: Inclined Planes

 

[asteriatic]

Can someone please explain the correct equation to use?


The normal force on an inclined plane is equal to the weight of the object perpendicular to the plane. In this case, the weight (mg) is acting downwards and the normal force is acting perpendicular to the plane, which means it is in the opposite direction of the weight. Therefore, the correct equation to use is Fn = mgcos30, as the cosine function represents the adjacent side of a right triangle, which is the side perpendicular to the incline. This will give the normal force as 5kg * 9.8m/s^2 * cos30 = 42.5N. It is important to remember that the normal force will always be less than the weight of the object on an inclined plane, as the weight is distributed over both the normal force and the parallel force (friction). I hope this helps clarify the concept for you.