How Do Horizontal Components Affect Total Tension in a Spring Balance?

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The discussion revolves around calculating the total tension in a spring balance affected by horizontal components of forces from two strings at angles α and β. The key equation established is T1*cos(α) + T2*cos(β) = Fb, where Fb represents the total tension measured by the spring balance. It is clarified that the weight W hanging from the system is balanced by the vertical components of tension, leading to T = W/sin(β) for the lower pulley. The conclusion drawn is that the total tension in the spring balance can be expressed as W*cos(α) + W*cos(β) = Fb, confirming the relationship between the angles and the forces involved. Understanding these components is essential for solving problems related to tension in pulley systems.
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Homework Statement


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.

Homework Equations


T - mg = ma
T1x + T2x = Fb


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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Forces on each side
of ideal pulleys are the same. Draw free body diagrams of each pulley, starting with the lower one.
 
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.
 
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?
 
Member69383 said:
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?
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).
 
For the FBD, it's just tension pulling up and weight (mg) pulling down. T = W?
 
So W*cos(α) + W*cos(β) = Fb?
 
Member69383 said:
So W*cos(α) + W*cos(β) = Fb?
If W is the weight, yes. If W is the mass then use Wg.
 
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Thanks, everyone =)
 
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