# Is the normal force always equal to the weight on an inclined ramp?

As the title states, my question is: Is the normal force always equal to the force of weight on an inclined ramp?

As an extension of this, when we are solving problems involving "a block slides down a frictionless, inclined plane that makes a 30 degree angle with the horizontal", why don't we have to take into consideration $$mg\cos{\theta}$$ or $$mg$$? I am told to solve it like this: $$a=\frac{F}{m}=\frac{mg\sin{\theta}}{m}=g\sin\theta$$

• Jay, are you learning physics on your own, or are you taking a physics class that includes an instructor? – David White Jun 6 at 3:43
• To pick up a point buried in Bob D’s answer: Whenever you face a problem like this, always consider the extreme cases. “An inclined plane” includes horizontal planes and vertical ones, and making a mental picture of those can clarify your intuition. This principle applies throughout physics. – Martin Kochanski Jun 6 at 4:30
• @DavidWhite I am learning on my own. – Jay Jun 14 at 23:35

The weight of the object, $$mg$$, is split into components down the ramp and normal to the ramp. These components are $$mg\sin{\theta}$$ and $$mg\cos{\theta}$$ respectively. So to directly answer your question, the normal force is never equal to the weight of the object on an inclined plane (unless you count the limiting case of level ground). It is equal to the weight of the object times the cosine of the angle the inclined plane makes with horizontal.

When computing the acceleration of an object down a frictionless inclined plane, we are only interested in the component of force (weight) down the plane, namely $$mg\sin{\theta}$$. Since the plane is frictionless, there is no contribution whatsoever from the normal force.

See here to visualize how the weight of the object is split into components:

As the title states, my question is: Is the normal force always equal to the force of weight on an inclined ramp?

No. The normal force equals the weight times the cosine of the angle the plane makes with the horizontal. If the angle was 90 dg (a vertical wall) the normal force would be zero and the mass would be in free fall with acceleration of $$g$$.. If the angle was 0 deg, the normal force would simply be the total weight since all of the weight would be supported by the surface.

I am told to solve it like this: $$𝑎=\frac{F}{m}=\frac{mgsin𝜃}{m}=gsin𝜃$$

Yes. because the force down (parallel) the plane is $$mgsin𝜃$$ whereas the force perpendicular (normal) to the plane is $$mg\cos𝜃$$. The only force that causes acceleration is the force down the plane, or $$gsin𝜃$$.

Hope this helps;

• @Chet Miller Thanks for the edit Chet – Bob D Jun 7 at 16:33