Centripetal and centrifugal forces

[adjacent]

Homework Statement

A space station is connected to a box(50kg) with a super strong string(1m) and it is rotating which makes the box rotate at a speed of 10m/s.
I am inside the box.
Find the force the box exerts on me.

Homework Equations

$F=m\frac{v^2}{r}$

The Attempt at a Solution

I found that centripetal force=5000N.
This force is exerted on the box by the string.
According to my experience,the "me" in the box would try to move opposite to the centripetal force direction but I heard that centrifugal force is not a real force.
So is the force exerted on me 5000N or something else?Can you explain why Newton's third law does not apply here?

 

[amind]

When we say centrifugal and coriolis forces are not real we mean they are virtual and they only exist if you are in a non-inertial reference frame,they are necessary to describe the motion in a non-inertial reference frame and you do experience it.

In your given problem , "you" will experience a force mv^2/r along the radius i.e. radially outwards

Now when you choose a non-inertial reference frame i.e. your reference frame itself is accelerating , you have to account for that acceleration too, and with your perspective the object you observe will also have an apparent acceleration due to your reference frame's acceleration.
This is why when you are rotating ( which is an accelerated motion ) you have to introduce these virtual forces.

 

[adjacent]

Oh.I understand now.
I have another similar problem:
The person A is inside a box and he is being accelerated at 10m/s^2 upwards.There are no external forces(Gravity,air resistance etc.)
A's mass is 50kg and the box is 1kg.
So according to Newton's second law,the upward force is 510N.
So,will I experience that 510N downwards?
I don't think so.If he exerts a 510N downwards and the rocket exerts a 510N upwards,the box would no longer be accelerating

 

[haruspex]

I found that centripetal force=5000N.
This force is exerted on the box by the string.

That would be true if the box were empty, but the question asks for the force the box exerts on you. Why would that be 5000N?

Can you explain why Newton's third law does not apply here?

It does apply. In an inertial reference frame, the force on you is unbalanced and results in an acceleration towards the spaceship. In your own frame, you are not accelerating but feel the force from the box, hence you invent centrifugal force to balance it.

 

[adjacent]

That would be true if the box were empty, but the question asks for the force the box exerts on you. Why would that be 5000N?

I can't find any other way to find a different value for the force

 

[Andrew Mason]

Oh.I understand now.
I have another similar problem:
The person A is inside a box and he is being accelerated at 10m/s^2 upwards.There are no external forces(Gravity,air resistance etc.)
A's mass is 50kg and the box is 1kg.
So according to Newton's second law,the upward force is 510N.
So,will I experience that 510N downwards?
I don't think so.If he exerts a 510N downwards and the rocket exerts a 510N upwards,the box would no longer be accelerating

You will exert a force on the box of 50 x 10 = 500 N. The box exerts a force on you of 500 N. The total force on the box is the applied external force of 510 N minus the 500 N that you are exerting, ie. 10 N , so the box accelerates at 10 m/sec^2.

AM

 

[adjacent]

You will exert a force on the box of 50 x 10 = 500 N. The box exerts a force on you of 500 N. The total force on the box is the applied external force of 510 N minus the 500 N that you are exerting, ie. 10 N , so the box accelerates at 10 m/sec^2.

AM

Thank you so much Andrew! This makes sense now
but:

I can't find any other way to find a different value for the force

 

[haruspex]

You will exert a force on the box of 50 x 10 = 500 N. The box exerts a force on you of 500 N. The total force on the box is the applied external force of 510 N minus the 500 N that you are exerting, ie. 10 N , so the box accelerates at 10 m/sec^2.

AM

Andrew, please reread the question. 50kg is the mass of the box, and has nothing to do with the force the box exerts on anyone inside it. Adjacent, to find the force it asks for you will need to weigh yourself.
Also, the speed of the box is 10m/s and the string is 1m long, so v²/r = 100, not 10. I don 't know where you got the 10N from.

 

[Andrew Mason]

Andrew, please reread the question. 50kg is the mass of the box, and has nothing to do with the force the box exerts on anyone inside it.

I was responding the post #3 which I quoted. In that scenario, the A is 50 kg. and the box is 1 kg.

AM

 

[haruspex]

I was responding the post #3 which I quoted. In that scenario, the A is 50 kg. and the box is 1 kg.

AM

Apologies.