Mechanics question?

Hi, I'm a little stuck on a Third Law problem.

It is as follows:

'A truck driver loads two identical untethered crates stacked one upon the other, and no sliding takes place. Each crate has 250 kg, therefore calculate the frictional force exerted on:

1) The upper crate by the lower
2) The lower crate by the deck of the truck

When the truck is accelerating at 1.5 ms^-2





What I've understood so far is that the truck would accelerate forward by 1.5^ms-2, which would then naturally cause both crates to move backwards by 1.5*250=375 N each?

But to counteract this and for them not to slide means there must be a frictional force acting in the direction of the truck. So I calculate that on the top this would be 375 N, but I don't know how to do the rest.

Thanks.

I would have said this is primarily Newton's 2nd law, rather than third.

For the top crate:

Since there is no slippage, the top crate accelerates at 1.5ms^-2, and by F=ma, the force acting on it is 375N in the direction the truck is moving.

THEN, by Newton's third law, there is a force acting in the opposite direction, of 375N, on the lower crate.

Now consider the lower crate:

Again Newton's second law gives the resultant force acting on the crate.

And this is the vector sum of the force the truck exerts and the force the top crate exerts.

Thank you.

May I ask another question?

It is as follows:

A man of mass 75 kg travels upwards in a lift which is decelerating at 0.4 m/s^2. Find the force exerted by the man on the lift.


I found the answer to be 705 N, which is equal to the reaction force. But what I don't understand in this case is that if the man exerts 705N on the floor, and the floor exerts 705N on him, how is the man exerting less than his weight*g on the floor?

Thank you.

May I ask another question?

It is as follows:

A man of mass 75 kg travels upwards in a lift which is decelerating at 0.4 m/s^2. Find the force exerted by the man on the lift.


I found the answer to be 705 N, which is equal to the reaction force. But what I don't understand in this case is that if the man exerts 705N on the floor, and the floor exerts 705N on him, how is the man exerting less than his weight*g on the floor?

I can't make sense of how the final part of your sentence follows from the bit in bold.

Don't know if this covers it.
If you've ever stood in a lift, going up, you will notice that as you start off, you'll feel heavier than normal, then as it reaches it's running speed, you'll feel your normal weight, and then as it comes to a stop you'll momentarily feel lighter.

If the force on the ground of the lift by the man = his reaction force, how come there is a net force downwards?

I would have said this is primarily Newton's 2nd law, rather than third.

For the top crate:

Since there is no slippage, the top crate accelerates at 1.5ms^-2, and by F=ma, the force acting on it is 375N in the direction the truck is moving.

THEN, by Newton's third law, there is a force acting in the opposite direction, of 375N, on the lower crate.

Now consider the lower crate:

Again Newton's second law gives the resultant force acting on the crate.

And this is the vector sum of the force the truck exerts and the force the top crate exerts.

i think you can treat the two crates as a single item thus ignoring the internal forces between them...