Homework Help: Newton's 2nd law - Tension

1. Feb 11, 2010

mybrohshi5

1. The problem statement, all variables and given/known data

A 5.4 kg mass and a 6.2 kg mass are tied to a light string and hung over a frictionless pulley. What is the tension in the string?

2. Relevant equations

F=ma

3. The attempt at a solution

Not sure how to do this one.

I know the tension on the string will be the same at both ends but not sure how to get the tension.

i am studyin for a quiz a have tomorrow and this is just a practice problem i found on a website.

could someone show me what to do on this one?

Thank you

2. Feb 11, 2010

jhae2.718

Create a separate free body diagram for each mass. There are two forces acting on each mass, the force of gravity, $F_g=mg$, and the tension. These forces act in opposite directions. Use a coordinate system such that the direction of motion is positive y.

Since the masses are not equal, there will be a net acceleration in the direction of the heavier mass. So, use ${\Sigma}F_y=ma_{net}$ to solve for a and then for T.

3. Feb 11, 2010

mybrohshi5

Ok i drew the FBD and i know that the heavier mass will acceleration downward and the lighter mass will therefore accelerate upward

but im not sure what you mean by --- what i quoted above??

4. Feb 11, 2010

mybrohshi5

would the net acceleration be 7.84 m/s^2

i got that from subtracting the masses then multiplying that by 9.8

5. Feb 11, 2010

jhae2.718

No. Here's a hint: $T-m_1g=m_1a$ and $m_2a=m_2g-T$, where m1 is the 5.4kg mass and m2 is the 6.2 kg mass.

Last edited: Feb 11, 2010
6. Feb 11, 2010

mybrohshi5

That didnt work out right for me. i came up with T=31.16

I solved for a from the first equation

a=(T-52.9)/5.4

then i plugged that into the other equation and solved for T

6.2((T-52.9)/5.4) = 6.2 - T

6.2T/5.4 + T = 6.2 - 328/5.4

T = 31.16

Can you see where i went wrong?

Thanks for the help

7. Feb 11, 2010

jhae2.718

You forgot to multiply m2 by g. See bold.

8. Feb 11, 2010

mybrohshi5

thank you :)

9. Feb 11, 2010

jhae2.718

Just to make sure: do you see how the expressions for tension can be derived by using the separate free body diagrams and setting the positive directions so the the direction of motion is always positive?

10. Feb 11, 2010

mybrohshi5

I kind of understand it and see where the equations come from but if you have a simple and easy way of explaining how to come up with those equations that would be very helpful for me.

thank you for the help

11. Feb 11, 2010

jhae2.718

The first step is always to draw a free body diagram. If there are multiple objects, create separate diagrams for each. Make sure to list all of the contact forces on each object.

When you have moving objects, it's best to use a coordinate system where the motion is always in the same direction. In this case, we have the downward direction for the larger mass be positive and the lower direction be positive for the smaller mass. This has the benefit of letting the system have a common net acceleration.

We then add up the forces in the FBDs and equate them with manet with accordance with Newton's second law. From here, in problems such as this we should be able to solve for the unknowns.

12. Feb 11, 2010

mybrohshi5

thank you. that helps a lot and makes perfect sense :)

13. Feb 11, 2010

jhae2.718

You're welcome. Not a problem.