# Force of hand on block = F of block on hand, not adding up?

1. Sep 11, 2015

### Ocata

Lets say I apply a force of 20N to a 10kg block on a frictionless table. Then the acceleration of the block is F= ma = F/m = 2m/s^2 = a.

Then, supposing I am standing on the ground with real world friction such that I am not accelerated in the opposite direction when I push the block that is on the frictionless table.

How do I calculate how much force the block applies to my hand as I push it?

If I assume the block pushes against my hand with the same amount of force that my hand applies to the block, then F = ma => 20N = m(2m/s^2) => 20N/(2m/s^2) = 10kg. So does that mean my hand weighs 10kg?

If I assume my hand weighs .5kg, then F = ma => F = .5kg*(2m/s^2) = 1N, which means the block is applying less force to my hand than my hand is applying to it, which I thought was not possible.

So what's going on here? Thanks

2. Sep 11, 2015

### DEvens

The hand bone connects to the arm bone...

You push on the block by moving muscles in your body. So the force from the block on your hand is, to some extent, balanced by the rest of you. Which in turn is to some extent balanced by the floor.

3. Sep 11, 2015

### Staff: Mentor

Be careful here. The F in F = ma is the net force. So the net force on your hand is 1 N. That, combined with the fact that the block exerts 20 N, allows you to determine that the force from the wrist is 19 N.

Edit: or is it 21 N? I'm not sure of the directions here.

4. Sep 11, 2015

### Staff: Mentor

When you apply F = ma to an object, be it the block or your hand, you must use the net force. (Better to write it as ΣF = ma.)

In this case, the 20 N force you exert on the block is the net force on the block, so you can calculate its acceleration easily.

Newton's third law tells you that the block exerts an equal and opposite force back on you, but that is not the only force acting on you or your hand. So that 20 N is not the net force on you or your hand.

5. Sep 11, 2015

### nasu

Adding to what was said already, your hand does not need to have the same acceleration as the block. It seems that you assume this in your OP when calculate the mass of the hand to be 10 kg.
Just by knowing the force of the block on your hand you cannot calculate your acceleration. Or of parts of your body.

6. Sep 12, 2015

### Ocata

nasu,

If my hand and the block do not have the same acceleration, then my hand is not applying a force to the block at all because there will be no contact. At different accelerations, there will be different velocities and therefor different positions of the hand and block, no? How can my hand apply a force if it is not in contact with the block? How can my hand maintain contact with the block if it is not contacting at the same position? How could they be in the same position if they are accelerating at different rates?

7. Sep 12, 2015

### nasu

Your hand may have the same acceleration as the block. I said it does not need to have it not that it cannot have it.
However, assuming that this acceleration of the hand is due to the force from the block, as you wrote in your post, is not justified.
If anything, that force will accelerate the hand in the opposite direction so it won't help maintaining the contact.
You need some other force acting on the hand. As it was already mentioned in the previous posts.

8. Sep 13, 2015

### Ocata

Okay, I'm getting it. The force from my hand onto the block is in fact equal to the force of the block on my hand, I just need to account for more components in this system.

9. Sep 13, 2015

### Ocata

Okay, I think I see what you are saying. If I try to think about it in terms of a free body diagram, the block has 20N applied to it with no friction so the block's net force is equal to the force applied to it. Applied Force = 20N, Opposing force = 0N, Net Force = 20N - 0N = 20N

Then the acceleration of the block is F/m = 2m/s^2 as previously calculated in the original post.

And given that the average human hand is roughly .5kg (Nasa's website says the average is 20 ounces), then, Fnet = ma = .5(2) = 1N

The mistake I was making was attributing 1N to the force applied onto the hand when it is actually the Net Force on the hand.

And since we know that 20N is applied to the Block by the hand, then 20N must be applied to the hand by the Block.

So [Fnet = Applied Force - Opposing Force] ==> [1N = 20N - X] = X = 19N

The Force that my wrist applies to my hand is 19N, but wait..

If I think about it in terms of a free body diagram, then it would mean that the block is actually applying more force than my wrist is applying to my hand. Thus my hand would travel in the opposite direction than I'm intending to describe..

So if I consider my wrist applying the greater force so that the net force is in the direction that I'm pushing, then my hand and the block will move in that direction and the net force on my hand will maintain a value of Fnet = 1N.

Thank you, very much, appreciate your clarification.