# 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.

To verify this, first weigh yourself on a weighing scale. Now, sit on it. Lean back and stretch your legs to counterbalance yourself, without touching the ground. Your weight will not have changed (you might need someone else to check the reading). If you have a longish object at home, you can try it with that, too (a plank, a rod, etc).

Note:The pressure exerted will depend on the areas, and so will the deformation of the smaller object. The pressure will be $\frac{mg}{\text{area of lower block}}$

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It makes so much obvious sense when I read your answer back to myself! I'm looking for more complex explanations than are necessary, I think. –  James Feb 10 '12 at 14:50
Don't worry, I've made the same mistake many times. Thinking too much is bad, too. =D –  Manishearth Feb 10 '12 at 14:55