Calculating Tension in Suspended Ball with Angles

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SUMMARY

The discussion focuses on calculating the tension forces (T1 and T2) in a suspended ball system with a weight of 5 Newtons. Participants emphasize the importance of using trigonometric functions, specifically sine and cosine, to resolve the tension forces into their horizontal and vertical components. The equations T1 + T2 = mg and T1 - T2 = 0 are established as essential for solving the problem, where mg represents the weight of the ball. A Free Body Diagram (FBD) is recommended as a crucial first step in visualizing the forces acting on the ball.

PREREQUISITES
  • Understanding of tension forces in physics
  • Knowledge of trigonometric functions (sine and cosine)
  • Ability to draw and interpret Free Body Diagrams (FBD)
  • Familiarity with vector resolution techniques
NEXT STEPS
  • Learn how to draw and analyze Free Body Diagrams (FBD)
  • Study the principles of vector resolution in physics
  • Explore the application of trigonometric functions in force calculations
  • Practice solving tension problems involving multiple forces and angles
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Students studying physics, particularly those focusing on mechanics, as well as educators seeking to enhance their teaching methods in force analysis and vector resolution.

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Homework Statement


A ball of weight 5 Newtons is suspended by two strings as shown above. Determine the magnitude of each of the forces, T1 and T2. (Note: sin 37° = 0.6; cos37° = 0.8)

t7hpvk.jpg



Homework Equations


T = mg


The Attempt at a Solution


I couldn't go anywhere because I need a formula that uses sin or cos, and I can't remember any right now. Can anyone supply me with the equations and a start? And I'll attempt at the solution with that. Please.
 
Last edited:
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Would this work? T x cos θ1 + T x cos θ2 = mg, where θ1 and θ2 are the angles the two sides.
 
Maty said:

Homework Statement


A ball of weight 5 Newtons is suspended by two strings as shown above. Determine the magnitude of each of the forces, T1 and T2. (Note: sin 37° = 0.6; cos37° = 0.8)

http://bb.mivu.org/courses/1/M-APhB1-S112-01/assessment/685f98ac2d604bddbf30b5bb28f9f01b/Q5.jpg


Homework Equations


T = mg


The Attempt at a Solution


I couldn't go anywhere because I need a formula that uses sin or cos, and I can't remember any right now. Can anyone supply me with the equations and a start? And I'll attempt at the solution with that. Please.

Maty said:
Would this work? T x cos θ1 + T x cos θ2 = mg, where θ1 and θ2 are the angles the two sides.

I'm having trouble accessing your image. Can you try uploading it to the PF?
 
Oops, there I edited the first post and it should be there now.

Also, after I posted that second post, I realized that nowhere in there it had any variables for the length of the string, which likely play a part in here. So I think I'm not supposed to use that equation.
 
Maty said:
Oops, there I edited the first post and it should be there now.

Also, after I posted that second post, I realized that nowhere in there it had any variables for the length of the string, which likely play a part in here. So I think I'm not supposed to use that equation.

On these problems, always start by drawing a Free Body Diagram (FBD) of the object (the mass in this case). Show the forces on the object, including the force down due to gravity. Then write the two equations to sum all of the forces in the x and y directions to zero (since the object is not moving). then solve away!
 
I have one drawn right in front of me right now, but I just don't know what formula to use, or how to find ALL of the forces in both axis.
 
Maty said:
I have one drawn right in front of me right now, but I just don't know what formula to use, or how to find ALL of the forces in both axis.

How many forces act in the horizontal direction? What are they? Label forces in the +x direction positive, and forces in the -x direction negative.

How many forces act in the vertical direction? What are they? Forces up are positive, and forces down are negative.

Since the object is not accelerating, you know that the sum of the forces in the x direction needs to equal zero. Same thing for the y direction. Those are the two equations you write and solve for the tension forces.
 
In the x direction T1 - T2 = 0?

And in the y direction. T1 + T2 = mg?

So i'll have to use Trig to find the numbers for each axis, but the only number I have is length, how would that help me?
 
Maty said:
In the x direction T1 - T2 = 0?

And in the y direction. T1 + T2 = mg?

So i'll have to use Trig to find the numbers for each axis, but the only number I have is length, how would that help me?

That is much closer. Good.

But you need to use the "components" of the tensions in the x and y directions, and that is where the sin and cos functions come in. Those are used to resolve vectors (like the tension forces applied at angles) into their x and y components.

See the "Intro to Vector Mathematics" link at the bottom of this FBD tutorial, for example:

http://physics.about.com/od/toolsofthetrade/qt/freebodydiagram.htm

.
 
  • #10
I am still back to the beginning. I know 'what' to do, I just don't know 'how' to do it. What numbers do I use, what equation?
 
  • #11
Your first step is to break up the tensions into vertical and horizontal parts, then since the ball isn't moving the vertical parts add to balance the force of gravity and the two horizontal parts must cancel with each other :)
 

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