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# Inclined plane force components

Explore the various forces acting on a block sitting on an inclined plane. Learn how to break the force of gravity into two components - one perpendicular to the ramp and one parallel to the ramp. Finally, using geometry and trigonometry, learn how to calculate the magnitude of each component of force that is acting on the block. Created by Sal Khan.

## Want to join the conversation?

• How can something be perpendicular to the surface of the earth which is curved. •   on a small scale it can be neglected.
if you stand on ground, you dont see it curved, you see it plane and straight. Thats because you are tiny comparatively
• Sal said normal force is perpendicular to the surface. But:

1. if normal force is reaction force (weight is 'action force') shouldn't it be positioned on a same line, that is, at same angle as weight of the body on a ramp? Third law states: "if one object exerts a force on another, the other exerts an equal and opposite force on the first." So, why is normal force 'normal' (perpendicular to the surface)?

2. In my experience, if a body is at inclined surface the part of the body closer to the angle of incline plane, (that is, left side of the box) pushes 'harder' to the ground than it's right side. If that is so, normal force shouldn't, in my opinion, be perpendicular to the surface because 'distribution of the pressure' on the inclined plane isn't even.

Thank you. Hope my question is clear... • If there is no friction the surface of on abject can only impart a force on another object perpendicular to surface, any component of the force parallel to the surface requires friction. So the force is broken up into a Normal Force and a Friction Force.

With an incline that is frictionless and you have a block on it the block's weight is directed strait down but the normal force is perpendicular to the incline so when you add the force vectors you end up with a net force parallel to the incline pointing down and this is what causes the block to slide down the incline.

On a frictionless incline the force is the same along the entire surface of the box so there is no portion that pushes harder into the ground than another. Once you introduce friction you have gravity and the normal force essentially acting like it is all at the center of gravity of the box but the friction force is along the surface of contact between the box and the incline. This friction force not in line with the center of gravity of the box produces a torque causing the edge of the box away from the direction of the friction force to put a bit more pressure on the incline that the other.

But regardless of the amount of pressure at each point along the the surface of contact the force without friction is always perpendicular to the surface and is the normal force.
• how do i know what trig-ratio to use??
I'm confused on that.. • What are the arrows that Sal is drawing above the F and G? • what is force
(1 vote) • i still dont understand why you break up the two forces of gravity. what's the point of doing this? • How would you calculate normal force for a box on an incline? And what about if there is a force being exerted horizontally which keeps the object at rest neglecting friction? How would to force diagram for a scenario like this look and how would you solve for Fnormal and Fapplied? • Breakout the gravity force vector into components which are parallel -mg*sin(Θ) and perpendicular -mg*cos(Θ) to the incline. The normal force will be equal and opposite to the perpendicular gravity component so N = +mg*cos(Θ). To keep the block from sliding, you would then need to apply a horizontal force equal and opposite to the parallel gravity component so that force would be F = +mg*sin(Θ).   