Basic physics - Forces of a block on a slope

andieje used Ask the Experts™

I asked a question a while back about the forces acting on a block on a slope. I only asked about the action force and not the normal force. Please could you look at the attached diagram and answer the questions about where exactly you draw these forces on the diagram.

Then it would be great if someone could tell me how to calculuate the normal force on the block.
I haven't put an angle of the slope in but I assume this will come into the calculation somewhere

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p.s. I've been told to assume at this basic level that the centre of mass and centre of gravity are one and the same thing


also, how do i know the line of the force F is parallel to the line of the slope? If it is it makes angle calculations easier
Rotate Rn by 180 deg so it is going into the block.

F (parallel to plane) is correct.

Also   Fn + Fp  =  mg
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>> Rotate Rn by 180 deg so it is going into the inclined plane.

Decompose  mg into components parallel and perpendicular (normal)  to the plane.
Draw the little right triangle using your mg vector as the hypotenuse.
You can move or redraw  Fp so that it starts at the CG of the block.
>>   >>   >> Rotate Fn by 180 deg so it is going into the inclined plane.



I thought the normal force was the force of the earth of the block. It obviously can't be from what you have said.

What have i misunderstood? Is the line that i drew equal to any force? Is it the reaction force?


Have you called it Fp for any reason


This is what my basic notes say:

The normal force is the support force exerted upon an object which is in contact with another stable object. For example, if a book is resting upon a surface, then the surface is exerting an upward force upon the book in order to support the weight of the book. On occasions, a normal force is exerted horizontally between two objects which are in contact with each other. For instance, if a person leans against a wall, the wall pushes horizontally on the person.

So the normal force should be exerted ON the block shouldn't it?
The gravitational force on the block can be decomposed into components parallel and perpendicular (normal)
to the direction of the inclined plane.   I was using the p for parallel.

If the block is to stay in contact with the plane, the normal to the plane must
be counteracted by force on the block from the plane ( reaction force or static support).

If the block is not sliding, the component of the gravitational force parallel to the plane must be canceled by
a friction force.

You original drawing was correct for the normal support force.
But to find the magnitude of that force, you have to do the vector decomposition in my drawing.
You can also find Fp from the decomposition, but you can't tell if it is large enough to overcome friction.
There is a component of gravitational force normal to the plane.
This force must be balanced/canceled by a normal force on the block from the plane.


Is this right then?
Correct if the block is not moving except 100/sqrt(2) = 70.71 ==>  should round up to 71


many thanks

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