Solving for Tension and Force in a Pulley System

  • Thread starter kxk010
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In summary, the problem involves a 2.0 kg ball tied to a string fixed to the ceiling and pulled by a force F. The force F is equal to the x component of the tension, which is also equal to the product of the tension and the sine of the angle between the string and the ceiling. The y component of the tension is equal to the weight of the ball. To solve for the force F, the equations for the sum of forces in the x and y directions are set to 0 and solved simultaneously. One method is to divide one equation by the other, eliminating the variable T and allowing for the calculation of F.
  • #1
kxk010
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First time taking physics in many years, so please forgive the simplistic nature of this problem.

A 2.0 kg ball tied to a string fixed to the ceiling is pulled to one side by a force F. Just before the ball is released, (a) how large is the force F that is holding the ball in position and (b) what is the tension in the string?

Also given: the angle between the string and the ceiling is 30 degrees.

I understand everything up to the point that I've found the components of T and calculated:

sum of x forces = Tcos30 (since in eqb sum of x forces=0)
sum of y forces = Tsin30 (since in eqb sum of y forces =0)

In the solutions manual, they are then taking F/w (force vector in positive-x direction/ weight vector in negative-y direction) = Tcos30/Tsin30

I don't understand why they are dividing F by w. I see that this will eliminate T so that you can solve for F, however, the decision to divide F by w seems arbitrary. Either that or I'm fundamentally missing something. Any insight?
 
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  • #2
kxk010 said:
sum of x forces = Tcos30 (since in eqb sum of x forces=0)
sum of y forces = Tsin30 (since in eqb sum of y forces =0)
What you have are not the sum of forces, but just the x and y components of the tension.

What other forces act on the ball?
Write equations for ƩFx = 0 & ƩFy = 0.

(I presume that the given force F is horizontal?)
 
  • #3
Doc Al said:
What you have are not the sum of forces, but just the x and y components of the tension.

What other forces act on the ball?
Write equations for ƩFx = 0 & ƩFy = 0.

(I presume that the given force F is horizontal?)

Yes, the force F is horizontal.

ƩFx = F - Tx = 0
so F = Tx
since Tx = Tsin30, F=Tsin30

ƩFy= Ty - W = 0
so Ty - w = 0
Ty = w
Tsin30 = w

So, in this instance, it turns out that the x and y components of the tension force are equal to F and W, respectively. So that's all I should be using to solve, right? Confused on the next step from here.
 
  • #4
What we have here is this

c3be4894.jpg


now all forces must cancel each other

so u get 2 equations and then solve for forceTension can be found similarly

Tsin30 = mg...equation (1)
F= Tcos30...equation (2)

m is given so find tension using equation 1
then put that value in equation 2 to get force
 
Last edited:
  • #5
Welcome to the forum kxk010!
You need to solve for F. The problem is of course that you don't know T so you have two unknowns.
You have to figure out a way to get rid of one of the two variables (T) .One way to go is to divide one by the other, as your solutions manual did, a substitution would also be fine.
 
  • #6
Edit.
 
  • #7
naptor said:
Welcome to the forum kxk010!
You need to solve for F. The problem is of course that you don't know T so you have two unknowns.
You have to figure out a way to get rid of one of the two variables (T) .One way to go is to divide one by the other, as your solutions manual did, a substitution would also be fine.

Ah, ok. So really you can take these 2 equations that you've "generated" and use them in any way necessary to isolate and solve for variables.
 
  • #8
kxk010 said:
Yes, the force F is horizontal.

ƩFx = F - Tx = 0
so F = Tx
since Tx = Tsin30, F=Tsin30

ƩFy= Ty - W = 0
so Ty - w = 0
Ty = w
Tsin30 = w

So, in this instance, it turns out that the x and y components of the tension force are equal to F and W, respectively. So that's all I should be using to solve, right?
Is the 30 degrees the angle between the string and the ceiling, as you said before? If so, rethink those components--you have them reversed.
Confused on the next step from here.
Once you get those equations fixed, you just have to solve them simultaneously for the unknowns T & F. There are several ways to do that--take your pick. One way, like they show in your solutions manual, is to just divide the two equations by each other. That will eliminate T and allow you to get F. Try it and see.
 
  • #9
Doc Al said:
Is the 30 degrees the angle between the string and the ceiling, as you said before? If so, rethink those components--you have them reversed.

Once you get those equations fixed, you just have to solve them simultaneously for the unknowns T & F. There are several ways to do that--take your pick. One way, like they show in your solutions manual, is to just divide the two equations by each other. That will eliminate T and allow you to get F. Try it and see.

You're right, I mistyped and F=Tcos30. Sorry about that. I seem to be stuck in this rigid mindset from other science classes that there is a sequential set of steps to solve every problem. Definitely finding that's not the case and there are multiple options (within limits) for the same problem. Thanks for all the help!
 

FAQ: Solving for Tension and Force in a Pulley System

What is tension/force problem?

Tension/force problem refers to a physics problem that involves calculating the amount of force or tension in a system. This type of problem is commonly encountered in mechanics and can involve multiple objects connected by strings, ropes, or cables.

How do you solve a tension/force problem?

To solve a tension/force problem, you need to first identify all the forces acting on the objects in the system. Then, use Newton's laws of motion and apply the equations of motion to determine the unknown forces. It is important to draw a free body diagram and establish a coordinate system to help with the calculations.

What are the key factors that affect tension/force in a system?

The key factors that affect tension/force in a system are the weight of the objects, the angle and direction of the strings or cables, and the acceleration of the objects. Other factors such as friction and air resistance may also play a role in the calculation of tension/force.

Can tension/force be negative?

Yes, tension/force can be negative. This usually occurs when the direction of the force is opposite to the direction of the motion. Negative tension/force does not necessarily mean that the object is being pulled in the opposite direction, but rather that the net force is in the opposite direction of the motion.

How is tension/force related to energy?

Tension/force is related to energy in the sense that work is done when a force is applied to an object. In a tension/force problem, the tension or force acting on an object is often used to calculate the work done, which is a measure of energy. Tension/force can also be used to calculate the potential or kinetic energy of an object in a system.

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