# Will the normal force will be different if there is different area of contact?

Let us say 2 soild frustum of height h and having different radius $$r_1$$ and $$r_2$$ respectively where $$r_1\>> r_2$$ having mass same. There is one one frustum which is kept on ground vertically where $$r_2$$ is facing downwards and $$r_1$$ is facing upward and just like that there is another frustum kept just beside that where $$r_1$$ is facing downwards and $$r_2$$ is facing upwards. (As shown in figure). Here $$n_1$$ is the normal force by figure 1 and $$n_2$$ is normal force by figure 2 So the question that is it possible that normal force will act different in both figure? Because its really confusing. This is what I came up with.

Let $$N_1$$ normal force acting on figure 1 and $$N_2$$ on figure 2.

If we try to make free body diagram in both figure then we will notice that mg will act downwards in both the situation and since mass is same in both figure then :

$$N_1 = mg , N_2 = mg$$

$$N_1 = N_2$$

So by this we can say that normal force are equal.

But then i thought that is it possible that N2 is greater than N1. This is what I came with :

$$P = F/A , F = PA$$

$$F \propto A$$

Since in figure 2 area of frustum is more downwards, then force will be greater since it is directly proportional to area then by that I can say that normal force will act greater in second than first. I am not sure that if this is correct explanation or not because this might be wrong but its really confusing what will be the answer. And if I keep thinking then again I came up with another explanation by taking center of mass or by taking torque.

So how normal force will act in both situation?

## Case 3: N1 < N2

• You are correct with your first assumption that the forces are equal. What dies change is the force exerted at a point – Jaywalker Oct 19 '15 at 16:29

Your working assumes Pressure remains constant. It does not.

For the body to be in translational equilibrium, the normal force and the weight must be equal and opposite (ignoring centripetal force from the earth's rotation). Since the weight is the same in both bodies, and both bodies are in translational equilibrium, the two normal forces must be equal.

Up to:

$N_1 = mg , N_2 = mg$

$N_1 = N_2$

You are entirely right.

But then you start confusing yourself with:

$P = F/A , F = PA$

$F \propto A$

The force $F$ is proportional to the surface area $A$ but it is also proportional to the pressure $P$, hence $F = PA$.

So regards the Normal force as a reaction force to the weight, only $mg$ needs to be taken into account:

## $N_1=N_2=mg$.

In the line where you are writing $P=F/A$ ... here you have alrady taken $F$ as a constant. Because, here the pressure is coming(or generating) due to the gravitational force, which depends only on $m$ and $g$ . So if $m$ and $g$ are same then here $F$ is constant. So you can't write $F$ as a variable depending on area.