Calculating Tension in a System of Trolleys

[Say17]

Homework Statement

Two trolleys with masses of 3kg and 1kg are connected by a wire. A force of 50N is exerted on the 3kg trolley and a force of 10N on the 1kg trolley. Any friction is negligible.

How much tension exists in the thread between the trolleys?

Relevant Equations

F = (m1+m2)a

Hi all,

To find out the tension. First I have to find out the a.

1. step
F = ma
F = (m1+m2) *a
F = (1+3)*a
F = 4*a
F/4 = a

F = 10N + 50 N
F = 60 N

60/4 = 15 m/s^2

2.step
T =( m1/ (m1 + m2)) * F
T = (1/ 1+3) * 60
T = 15 N

But the final result is 20N.

Thanks for your help!

 

[kuruman]

Homework Statement: Two trolleys with masses of 3kg and 1kg are connected by a wire. A force of 50N is exerted on the 3kg trolley and a force of 10N on the 1kg trolley. Any friction is negligible.

How much tension exists in the thread between the trolleys?
Relevant Equations: F = (m1+m2)a

F = 10N + 50 N

The two forces act in opposite directions, therefore the net force on the system of two trolleys must be less than the sum of their magnitudes.

 

[Say17]

So it means one of them is minus?

 

[Mister T]

If you pull to the right with 50 N of force, and your friend pulls to the left with 10 N of force, what is the net force?

Also, where are you getting the equation $T=\frac{m_1}{m_1+m_2}F$?

 

[Say17]

I saw this formula somewhere and thought this may work......
So it is 50N - 10N = 40N?

 

[kuruman]

I saw this formula somewhere and thought this may work......
So it is 50N - 10N = 40N?

Forget the formula you saw and reason it out.
If a force of 50 N pulled to the left and another 50 N force pulled to the right, what would be the net force on the two trolleys?
What would happen if you reduced the force to the left from 50 N to 10 N?
What would happen if you reduced the force to the left from 10 N to zero?

 

[Say17]

If a force of 50 N pulled to the left and another 50 N force pulled to the right, what would be the net force on the two trolleys? 0
What would happen if you reduced the force to the left from 50 N to 10 N? 40N
What would happen if you reduced the force to the left from 10 N to zero? 10 N

 

[kuruman]

If a force of 50 N pulled to the left and another 50 N force pulled to the right, what would be the net force on the two trolleys? 0
What would happen if you reduced the force to the left from 50 N to 10 N? 40N
What would happen if you reduced the force to the left from 10 N to zero? 10 N

Great! You answered your own question in post #5. Now you can finish the problem.

 

[Say17]

Then I have to divide 40 N by 2?

 

[kuruman]

Then I have to divide 40 N by 2?

Why 2? What does 2 represent?

Just modify the two steps that you showed originally in post #1 knowing what you know now.

 

[PeroK]

Why 2? What does 2 represent?

There are two trolleys. They get $20N$ each is the idea, I guess.

 

[Say17]

Yes exactly, because there are two trolleys.

 

[PeroK]

Yes exactly, because there are two trolleys.

Why would the force be split evenly between the two trolleys?

 

[Say17]

Why would the force be split evenly between the two trolleys?

Oh you are right. They have different mass.

 

[nasu]

Even if they had same mass, dividing the net force on the system by two has no meaning. The external force on one trolley is 50N and the external force on the other is 10N. The internal force on each is the tension in the string. You either write N's law for the system, using the net external force (40 N in magnitude) or you write it for each body with the actual forces acting on that body.

 

[MatinSAR]

First I have to find out the a.

Using $F_{net}=ma$ you can find acceleration of this system. Suppose that the length of the thread is constant, so both objects have the same acceleration.
You can find the net external force that acts on system with both objects:

You simply find net external force and mass of the system using above picture. then you can find acceleration of system which is equal to the acceleration of each object. After finding acceleration use $F_{net}=ma$ for one of the objects then you can find tension force. You need to add two more forces to the picture above then you can find that what is the net force($F_{net}$) that acts on each object.

 

[Mister T]

I saw this formula somewhere and thought this may work......

No, it's a formula for the tension in a cord connecting two objects in a particular situation. Not this situation, though. When you look at the two situations you can see that they are entirely different.

So it is 50N - 10N = 40N?

Yes, that's the net force on the two-object system. Use it to find the acceleration of that system. Then use that acceleration to find the net force on either one of the objects.

 

[kuruman]

You simply find net external force and mass of the system using above picture. then you can find acceleration of system which is equal to the acceleration of each object. After finding acceleration use $F_{net}=ma$ for one of the objects then you can find tension force. You need to add two more forces to the picture above then you can find that what is the net force($F_{net}$) that acts on each object.

Actually, when one has the common acceleration of the masses, one does not need to add two or more forces to find the net force on each object. Just multiply the mass of the object by the common acceleration.

 

[MatinSAR]

Actually, when one has the common acceleration of the masses, one does not need to add two or more forces to find the net force on each object. Just multiply the mass of the object by the common acceleration.

I meant that by adding other forces, we determine the net force on the object which has mass of 1 kg as $T-10N$ then then we substitute in N's law.
$F_{net}=ma$
$F_{net}=T-10N$
$T-10=ma=10$ so $T=20N.$