Tension in string in system with multiple objects?

[jb007]

Homework Statement

I am stuck on this problem:
So the question is asking for F₂₁, F₁₂, F₃₂, and F₂₃. (That's the force by 2 on 1).
The mass of m₂ is variable.
There are four cases where I need to solve for the forces: when m₂ has mass of 3kg, 0.3kg, 0.03kg, and 0kg.
The mass of m₁ = 2.5kg, the mass of ₃ = 3.5kg.
Also assume that friction is neglected.
I am confused on how to resolved the tension force in a system with numerous objects.
The T is constant, 3N. imagine there's a person pulling on the string with a constant force of 3N.
I know the acceleration of all three carts in each case should be the same.
But how do I solve for the tension force in the strings between the carts?

Homework Equations

F=ma
Newton's 3rd Law
Pretty sure that's all I really need here.

The Attempt at a Solution

[/B]Ok, so I drew free-body-diagrams for each case. I know T is always equal to 3N. The weight force and normal force of all three carts aren't taken into account.
Should the tension force in the strings be constant?

Suppose I were to do case 1 with m₂ = 3kg:
T=3N
F_net = 3N
3 = (m₁+m₂+m₃)a
a = 3/(2.5+3+3.5) = 0.33 m/s/s (for the whole system, so each cart has the same acceleration)

Going from left to right, solving for F₂₃, would the force m₂ on m₃ be equal to (m₂+m₁)0.33?
Then would F₃₂ be equal in magnitude but in the opposite direction of F₂₃ because of Newton's 3rd Law?

I'm not sure what to include or exclude when I'm calculating a force within a specific part of a system.

Any pointers would be very helpful. Thanks.

 

[Nathanael]

Suppose I were to do case 1 with m₂ = 3kg:
T=3N
F_net = 3N
3 = (m₁+m₂+m₃)a
a = 3/(2.5+3+3.5) = 0.33 m/s/s (for the whole system, so each cart has the same acceleration)
...
Any pointers would be very helpful. Thanks.

I suggest waiting until the end to plug in the values of m₂ :)

Should the tension force in the strings be constant?

Yes, because the acceleration is constant. But don't worry about that; it will come out of the math.

Then would F₃₂ be equal in magnitude but in the opposite direction of F₂₃ because of Newton's 3rd Law?

Yes this is true.

I'm not sure what to include or exclude when I'm calculating a force within a specific part of a system.

Okay let's just consider the cart m₃. For right now, m₃ is our entire system. What is the net force on the system? And how does it relate to acceleration? (Just write "a" for the acceleration for now)

Then treat m₂ as your entire system. Write a net force equation for it. Then do m₁

 

[jb007]

Thanks for the reply.
Treating each cart as its own system, I'm getting here: (to the right is positive force, to the left is negative force)

The net force acting on m₁ is T-F₂₁. Which is equal to 3-(m₂a)?
The net force acting on m₂ is F₁₂-F₃₂. Which is equal to (m₁a)-(m₃a)?
The net force acting on m₃ is just F₂₃, which is equal to m₂a?

And the acceleration is the same for all carts, so the a value should be F/(m₁+m₂+m₃)? Then is it just 3/(m₁+m₂+m₃)?

 

[haruspex]

The net force acting on m₁ is T-F₂₁.

Yes.

Which is equal to 3-(m₂a)?

T = 3 (N), but why should F₂₁ = m₂a?

The net force acting on m₂ is F₁₂-F₃₂.

Right, and net force produces acceleration, so according to ##\Sigma F = ma## that net force is equal to ...?

 

[jb007]

Yes.

T = 3 (N), but why should F₂₁ = m₂a?

Right, and net force produces acceleration, so according to ##\Sigma F = ma## that net force is equal to ...?

Ok, so for m₂, the F_net equation would be F₁₂-F₃₂ = m₂a₂
Then I could solve for a₂ by dividing F_net by m₂.
But how could you calculate the individual forces, F₁₂ and F₃₂?

 

[haruspex]

But how could you calculate the individual forces, F₁₂ and F₃₂?

You have three equations, ##\Sigma F = ma## for each of the three masses.
You have three unknowns: two forces and an acceleration.
Solve.

 

[jb007]

Ok, so for each case I can solve for the acceleration for the entire system.
Then I break up the system into each mass component and treat them independently.

The F_net of m₁ would be T-F₂₁ = m₁a

The F_net of m₂ would be F₁₂-F₃₂ = m₂a

The F_net of m₃ would be F₂₃ = m₃a

I first solve for F₂₁ by using the first equation, then I solve for F₃₂ using the second equation, then I know the F₂₃.

Solved it. Thanks all.