Solving for Tension and Force in a Pulley System

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The discussion focuses on solving for tension and force in a pulley system involving a 2.0 kg ball suspended by a string at a 30-degree angle. The key equations derived are ƩFx = F - Tcos30 = 0 and ƩFy = Tsin30 - W = 0. To find the force F, the participants clarify that dividing the equations allows for the elimination of tension T, leading to the relationship F/w = Tcos30/Tsin30. This method effectively isolates F, enabling a solution for the force acting on the ball.

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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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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?)
 
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.
 
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:
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.
 
Edit.
 
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.
 
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.
 
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!
 

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