Where would $mg$ act if I consider a man standing on a block and consider the block as my system?

When a man stands on a block, the weight of the man i.e. $$mg$$ acts downward on the block and Normal reaction $$N$$ in opposite direction (provided I consider the man as my system). However, if I consider the block to be my system, the normal reaction is now provided by the man standing on top of it. Where would $$mg$$ act in this case? In other words, what balances the action of the Normal Force (provided by the man)?

$$mg$$ acts on the man;

The man's feet act on the block; This force happens to be equal to $$mg$$;

The block acts on the man's feet; You called this force $$N$$.

Feet's action downward balances $$N$$.

• @J.Murray The block's weight does not matter here. Jul 8, 2021 at 12:40
• Oh, my mistake, I misread. I thought $N$ was referring to the normal force the floor exerts on the block. Sorry. Jul 8, 2021 at 12:42

The man is not exerting a normal reaction on the block. He is exerting his weight on the block, and the ground is exerting a normal reaction on the block that compensates the weight of the man and the block.

You wanted to consider the block as your system so here are the three forces acting on the block:

-the man weighing down on it

-the weight of the block itself

-normal reaction of the ground compensating both

• This isn't right. The man's weight is the gravitational force on him. The downward force he exerts on the block is equal in magnitude to the upward normal force which the block exerts on him. This will be equal to his weight if he's standing motionless on the block, but if he's in the process of jumping (or if he's holding something in his hands) then this will no longer be true. Jul 8, 2021 at 12:14
• Of course without specific indication from OP that the man is doing his workout, I assumed he wasn't... Jul 8, 2021 at 12:26
• My point is that "the man exerts his weight on the block" is incorrect. The force he exerts on the block may or may not equal his weight depending on the circumstances. If you stand on a scale and wave your arms, the reading on the scale will change. Would you say that your weight is changing, or that the force you're exerting on the scale is changing because you're moving around? As written, your answer would indicate the former. Jul 8, 2021 at 12:29
• I can agree that "the man exerts a force equal to his weight" would technically be a more accurate formulation for a man standing still (as the specifics of the question imply) Jul 8, 2021 at 12:44
• All I'll say is that I've taught mechanics a fair few times, and it is an extremely common misconception that the downward force on a surface is equal to the weight of the object which is directly in contact with it. It's not true if the object in question is supporting another one, it's not true if there is acceleration (e.g. the whole system is in an elevator), it's not true if part of the system is accelerating (e.g. the man is picking up a child), and so on. My belief is that it's easier to be careful at the beginning than to go back and be careful later, but that's just my opinion. Jul 8, 2021 at 12:56

Weight of any object acts on that particular object and its reaction acts on the centre of the earth. Here, if you consider only man as the system, the forces acting on him will be his weight and the normal reaction exerted by the block balanced by each other. If you consider the block as the system, those will be the block's weight, normal reaction exerted by the man and the normal reaction by the floor. If you consider both man and the block as the system, the normal forces on the man and the block exerted by each other will be cancelled as they become internal forces. Internal forces of a system do not affect the motion of it. So here you have weights of the man and the block acting downwards balanced by the normal force exerted by the floor acting upwards. Those three forces are the external forces in this situation.