# Why should the Normal apply in the downwards direction?

As part of a problem, I was required to make a free body diagram for the following system:

It is given that the platform and the man have the same mass $$m$$. Apart from the forces due to tension, my initial thought was that the platform would apply a normal force $$N$$ upwards due to the weight $$mg$$ of the man.

But in the solution to the problem, apparently, there is a normal force $$N$$ acting downwards. The normal is force is given by the man, to the platform. Why is that? Why is the normal force acting in the direction of weight?

For most other problems, I am used to applying the normal force upwards, usually given by a platform/wedge to the object on top of it, to balance out the downward force given by the object due to its weight. But in this case, the normal force is acting downwards and I cannot understand the reason why.

For most other problems, I am used to applying the normal force upwards, usually given by a platform/wedge to the object on top of it, to balance out the downward force given by the object due to its weight. But in this case, the normal force is acting downwards and I cannot understand the reason why.

You would consider the normal force exerted upwards by the platform on the person, if you were drawing the free body diagram of the man.

Here you have drawn the free body diagram of the platform. Hence, you must consider the normal force exerted downwards on the platform , as is shown in the diagram

• So here we show the normal force is acting downwards because the FBD is drawn from the perspective of the platform. Then, $N = mg$ (because of the weight $mg$ of the man). If we had drawn the FBD of the man, then we would have shown the normal to be acting upwards due to the platform and equal to the man's weight. Please correct me if I am wrong in my understanding. Commented Apr 27, 2021 at 9:06
• You are correct. Except for value of N. I am not sure if it would be equal to the weight of the man, because there is a vertical rope that would also have some tension T, that would affect, what is the net normal force exerted by man on platform and by platform on man Commented Apr 27, 2021 at 9:54
• Yes, I think the accurate force equation would be $T_{1} +T_{2} = N + mg$, where $T_{1}$ and $T_{2}$ are the tension forces due to the left and right rope respectively. Commented Apr 27, 2021 at 10:13
• No, for the FBD of the platform the equation would be T1 + T2 = N Commented Apr 27, 2021 at 10:16
• what about the mass of the platform itself? Commented Apr 27, 2021 at 10:18

There is a normal force on the man due to the platform in the upward direction.

The Newton third law pairing is the force, of equal magnitude, on the platform due to the man in the downward direction.