Newton's third law problem

[00945584]

Two crates are moving along a frictionless horizontal surface. The first crate, of mass M=100kg, is being pushed by a force of 300N. The first crate is in contact with the second crae, of mass m= 50kg.


1. What is the force exerted by the larger crate on the smaller crate?
2. What is the force exerted by the smaller crate on the larger crate?

I am confused on the matter of setting the force vectors such as F12 or F 1 on 2 or F 2 on 1 and getting the answer. Everytime I try a method to get the answer the way I set the force vectors makes no sense.


Any help would be appreciated. Thanks!!

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[loveoforganic]

So you have your 2 boxes - 100 kg and 50 kg. They're in contact, and there's no friction, so picture them as one giant box of mass 150 kg. You have a force of 300 N on this 150 kg box. F = ma. 300 / 150 = a = 2 m/s (for the entire system).

F = ma = (50 kg)*(2 m/s) = 100 N on the smaller box in the direction of motion.

F = ma = (100 kg)*(2 m/s) = 200 N on the larger box in the direction of motion.

 

[Chowdder]

That's actually not the answer he is looking for. This question is referring to a case of Newton's third law of motion where he is trying to determine the action/reaction forces exterted between the boxes. This is actually one of those questions that I dread for when I was taking first year physics, but I'll show you how I would go about solving it.

The answer you're looking for is 100N

First of all, you really need to draw out the problem as free-body diagram for each individual boxes. I am going to neglect the normal and the gravitational forces since they cancel each other out and do not contribute to the solution.

I'll set out the variables
F: force applied
F2on1: force of box2 on box1
F1on2: force of box1 on box2
m1: mass of box 1
m2: mass of box 2
a: acceleration

For box 1, since it is the box that is being pushed, you have the force applied at 300N to the direction of motion. You also have an opposing force coming from box 2. This force is F of box 2 on box 1, F2on1.

For box 2, it is the box that is being pushed by box 1. This box is not being directly pushed, but rather experience the pushing force indirectly through box 1. This force is F of box 1 on box 2, F1on2.

Now you can layout the formula through Newton's second law.
For box 1:
F - F2on1 = m1*a

For box 2:
F1on2 = m2*a

You can calculate a simply by assuming the both boxes are one single box. Then you have F=(m1+m2)*a
a turns out to be 2 m/s

Plug a into the equation and you get 100N for both F1on2 and F2on1. This makes sense because they have to equal to each other, otherwise one box will begin to accelerate faster than another.

 

[Cjc555777]

which one of you are correct? I thought I knew what I was doing, but now I'm confused...

 

[loveoforganic]

We're both right, but he's more right He was right when he said I failed to answer the question - didn't look closely enough. So for the full, correct answer, and not just part of the work to get to the right answer, refer to his post. If you wanted to expand on the work I did, you could see that the 300 N force applied to the first block had to be counteracted somehow to get reduced to 200 N net force. 300 N - 200 N leaves you with 100 N acting on the first block by the second block, which is the same force acting on the second block by the first block.