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

  • Thread starter Thread starter Member69383
  • Start date Start date
  • Tags Tags
    Angles
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
9 replies · 2K views
Member69383
Messages
7
Reaction score
0

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?
 

Attachments

  • Tension2.jpg
    Tension2.jpg
    12.8 KB · Views: 527
Physics news on Phys.org
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?
 
Thanks, everyone =)