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

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Homework Help Overview

The discussion revolves around a problem involving a spring balance measuring total tension in a string system with weights and pulleys. The original poster presents a scenario where two tensions, T1 and T2, are at angles α and β, respectively, and questions how the horizontal components of these tensions contribute to the reading on the spring balance.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between the tensions in the strings and the weight being supported. There are attempts to derive equations based on the horizontal components of the forces and discussions about the nature of tension in the system.

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions about the forces involved and discussing the implications of the angles on the tensions. Some guidance has been offered regarding the relationships between the forces, but there is no explicit consensus on the final formulation of the equations.

Contextual Notes

There are indications of potential confusion regarding the definitions of weight and mass, as well as the need for clarity on the setup of the free body diagrams. The discussion also highlights that all pulleys are anchored down, which may affect the tension calculations.

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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.
 
  • #10
Thanks, everyone =)
 

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