# Tension/Forces at Angles

1. Mar 2, 2013

### Member69383

1. The problem statement, all variables and given/known data
Fb is the spring balance which measures the total tension in the string. So T1 + T2 = Fb
The system is at rest so a = 0.
Weight W is hanging as shown in the diagram.
T1 corresponds with angle α and T2 corresponds with angle β.

ONLY THE HORIZONTAL FORCE CONTRIBUTES TO THE AMOUNT IN FB!

Find the horizontal components of the force exerted by each string.

2. Relevant equations
T - mg = ma
T1x + T2x = Fb

3. The attempt at a solution
I know that Force 1*cos(α) + Force 2*cos(β) = Fb

Is Force 1 = Force 2 = W?
Or is it Force 1 = Force 2 = 2W?

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2. Mar 2, 2013

### PhanthomJay

Forces on each side
of ideal pulleys are the same. Draw free body diagrams of each pulley, starting with the lower one.

3. Mar 2, 2013

### haruspex

First think about the string. If the tension varies along the string, what will the string do?
Next, figure out what the tension is in the bottom part of the string.

4. Mar 2, 2013

### Member69383

Upper pulley FBD = Tension to the left at angle α and I guess some type of support holding it up?

Lower pulley FBD = Tension to the left at angle β as well as weight pointing down.

For the lower one, it's supported by the Y component of that string, right?

T*sin(β) = W
T = W/sin(β)

If forces on each side are the same, then (W/sin(β))*cos(α) + (W/sin(β))*cos(β) = Fb?

5. Mar 2, 2013

### PhanthomJay

no. Although not shown , all pulleys are anchored down. If you look at the lower pulley, what is the tension in the section of the rope that is holding up the weight (draw an FBD of the weight).

6. Mar 2, 2013

### Member69383

For the FBD, it's just tension pulling up and weight (mg) pulling down. T = W?

7. Mar 2, 2013

Yes.

8. Mar 2, 2013

### Member69383

So W*cos(α) + W*cos(β) = Fb?

9. Mar 2, 2013

### haruspex

If W is the weight, yes. If W is the mass then use Wg.

10. Mar 2, 2013

### Member69383

Thanks, everyone =)