# Homework Help: Find the Tension

1. Feb 29, 2012

### Rey4312

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

In the figure below, assume that the slope is frictionless and that the two blocks are connected by a massless cord. Assume the following:
θ1 = 37°
θ2 = 45°
m1=3kg
m2=0.86kg.
What is the tension in the cord?.

2. Relevant equations

F=ma , w=mg, trig

3. The attempt at a solution

i know that you need to find acceleration. Which i did (a=3.04) and i found it by ((m1)(m2)g)/(m1+m2)

you'll have to use trig functions but i'm not sure which one to use.
And wouldn't you set it equal to each other?

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2. Feb 29, 2012

### Staff: Mentor

Where did that formula come from?

In any case, try analyzing the forces on each block separately. Draw a FBD for each and apply Newton's 2nd law. Then combine the two equations to solve for the acceleration and the tension.

3. Feb 29, 2012

### Rey4312

I was given it...

Could you use the trig function tangent?
so then your equations would be:
w tan(37)=3 a
and
w tan(45)=.86 a

but then you're only left with acceleration, and theres two different ones.

4. Feb 29, 2012

### Staff: Mentor

It doesn't apply to this problem.

What forces act on m1 parallel to the surface?

What forces act on m2 parallel to the surface?

5. Feb 29, 2012

### Rey4312

oh okay...

the only force that i can think of would be friction, at least that is parallel

6. Feb 29, 2012

### Staff: Mentor

No. You are told that the slope is frictionless.

There are two other forces with components parallel to slope. What are they?

7. Feb 29, 2012

### Rey4312

w=mg, tension?, or would it be the normal force (though i thought that was perpendicular)

8. Feb 29, 2012

### Staff: Mentor

Right, those are the two forces with parallel components. What is the component of the weight parallel to the slope?
Yes, the normal force is perpendicular. So you won't need it.

9. Feb 29, 2012

### Rey4312

W= mg (mass*gravity)

10. Feb 29, 2012

### Staff: Mentor

What direction does the weight act?

11. Feb 29, 2012

### Rey4312

directly down parallel with the slope

12. Feb 29, 2012

### Staff: Mentor

Directly down, yes. But that's not parallel to the slope. (The slope isn't vertical.) You need to find the component parallel to the slope.

You might want to read this: Inclined Planes

13. Feb 29, 2012

### Rey4312

the component parallel would be tension.

14. Feb 29, 2012

### Staff: Mentor

The tension in the string is a different force, but yes it is parallel to the slope. You still need the component of the weight parallel to the slope. (Read the link I gave.)

15. Feb 29, 2012

### Rey4312

Sorry didn't see that the first time..

so it would be net force?

16. Mar 1, 2012

### Rey4312

solved it... thanks for the help