## Find the Tension

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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 Quote by Rey4312 i know that you need to find acceleration. Which i did (a=3.04) and i found it by ((m1)(m2)g)/(m1+m2)
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.
 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.

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## Find the Tension

 Quote by Rey4312 I was given it...
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?

 Quote by Doc Al It doesn't apply to this problem.

oh okay...

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

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 Quote by Rey4312 the only force that i can think of would be friction, at least that is parallel
No. You are told that the slope is frictionless.

There are two other forces with components parallel to slope. What are they?
 w=mg, tension?, or would it be the normal force (though i thought that was perpendicular)

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 Quote by Rey4312 w=mg, tension?,
Right, those are the two forces with parallel components. What is the component of the weight parallel to the slope?
 or would it be the normal force (though i thought that was perpendicular)
Yes, the normal force is perpendicular. So you won't need it.
 W= mg (mass*gravity)

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 Quote by Rey4312 W= mg (mass*gravity)
What direction does the weight act?
 directly down parallel with the slope

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 Quote by Rey4312 directly down parallel with the slope
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
 the component parallel would be tension.

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 Quote by Rey4312 the component parallel would be tension.
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.)