# Calculating the force of a mass bearing down on an object

I need to calculate the force of a weight bearing down on a smaller object, constraining it from expansion.

The weight bearing down has a much bigger surface area than the smaller object.

I know the mass of the weight.

I know the surface area of the smaller object.

What else do I need to know in order to calculate the force bearing down on the small object?

What I mean is, what do I need to know in order to calculate what the force bearing down on that smaller object is?

I think I might need to know about the acceleration of gravity (9.8 m s^-2), but I'm not sure how that applies if I don't know the dimensions of the larger object.

I know that F=ma, but I can't visualise how that applies to an object bearing down on an object with a smaller surface area.

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Hi James - what makes you think you need to know something else? Have you tried looking for a formula that would allow you to calculate pressure? Does it suggest what you might need? – David Z Feb 10 '12 at 0:20
Hi David - I might have been a bit vague in my question. What I mean is, what do I need to know in order to calculate what the force bearing down on that smaller object is? I think I might need to know about the acceleration of gravity (9.8 m s^-2), but I'm not sure how that applies if I don't know the dimensions of the larger object. I know that F=ma, but I can't visualise how that applies to an object bearing down on an object with a smaller surface area. Unfortunately, my understanding of physics and maths is patchy, so I'm not clear on how some of the fundamentals relate to each other. – James Feb 10 '12 at 1:06
That would be a good explanation to include in your question. – David Z Feb 10 '12 at 1:51
I've updated the question with my comments from above. – James Feb 10 '12 at 2:21

The force is $mg$. Thats it. Area does not come into the picture here. Whats happening is this: At equilibrium, all accelerations are zero, so every force must be balanced. The large object has a wight of mg, so to balance that, the small object exerts the a normal force $N=mg$ back up on the large block. Every action has an equal and opposite reaction, so the upper block exerts the same force N=mg down on the block. Thats it.
Note:The pressure exerted will depend on the areas, and so will the deformation of the smaller object. The pressure will be $\frac{mg}{\text{area of lower block}}$