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

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In summary, the homework statement states that the total tension in a string is equal to the sum of the tensions in the two strings it is hanging from. The system is at rest, so a = 0. The horizontal force on each string only contributes to the amount in FB.
  • #1
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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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  • #2
Forces on each side
of ideal pulleys are the same. Draw free body diagrams of each pulley, starting with the lower one.
 
  • #3
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
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
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).
 
  • #6
For the FBD, it's just tension pulling up and weight (mg) pulling down. T = W?
 
  • #7
Yes.
 
  • #8
So W*cos(α) + W*cos(β) = Fb?
 
  • #9
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 =)
 

What is tension/forces at angles?

Tension/forces at angles refers to the forces or pulls acting on an object at an angle, rather than directly in line with the object.

What is the difference between tension and force?

Tension is a type of force that is transmitted through a rope, cable, or other similar object. It is often used to pull or support an object. Force, on the other hand, is a push or pull that can cause an object to accelerate or change its motion.

How do you calculate tension/forces at angles?

The calculation for tension/forces at angles involves using trigonometry, specifically the sine and cosine functions. You will need to know the magnitude of the force, the angle at which it is acting, and the components of the force in the x and y directions. The formula is Tension = Force * sin(angle).

What factors can affect tension/forces at angles?

The magnitude and direction of tension/forces at angles can be affected by a number of factors, including the angle at which the force is applied, the weight and shape of the object, and the material and strength of the rope or cable used to transmit the force.

Why is understanding tension/forces at angles important in science?

Understanding tension/forces at angles is important in science because it allows us to accurately predict and analyze the behavior of objects and structures that are subject to these types of forces. This knowledge is essential in fields such as engineering, physics, and mechanics.

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