Determining T and Theta in a 10kN Pulley System

[Ry122]

http://users.on.net/~rohanlal/Q5.jpg
The maximum force of the block, T, must be determined. The 2nd rope that the pulley is attached to has the same maximum tension force as the rope of the pulley which is 10kN.
Theta also needs to be determined.

 

[Dick]

Could you possibly write the sum of all the forces on the pulley and set it equal to zero? There are two unknown tensions in the problem, T and the unnamed tension between the pulley and the support point. Balancing the forces should give you two equations in two unknowns.

 

[Ry122]

is this what you mean?
make T = 10 since it will undergo the highest tension.
change in y = 10cos0-w=0
how do i do the x component?

 

[Dick]

I'm not even completely sure what your original question means. I was hoping if you started solving it I could figure that out, but I guess I was wrong. Why T=10? y=10cos0-w=0 looks kind of like a force balance equation for the weight. Is that what it's supposed to be? How about balancing the forces on the pulley?

 

[nrqed]

I'm not even completely sure what your original question means.

The way I interpret it is the following: There are three unknowns (the two tensions and the angle) and the constraint that neither tension must be over 10 kN.

The approach I would suggest would be to first assume that the tension T is the maximum value, 10 kN. And then solve for the other tension and the angle theta. If the other tension is less than 10 KN then that's it, that's the final answer to the problem.
If the other tension is above 10 kN, one has to start all over again by setting that other tension equal to 10kN and finding T and theta.

 

[Dick]

The way I interpret it is the following: There are three unknowns (the two tensions and the angle) and the constraint that neither tension must be over 10 kN.

The approach I would suggest would be to first assume that the tension T is the maximum value, 10 kN. And then solve for the other tension and the angle theta. If the other tension is less than 10 KN then that's it, that's the final answer to the problem.
If the other tension is above 10 kN, one has to start all over again by setting that other tension equal to 10kN and finding T and theta.

Ah! That makes sense, thanks.

 

[Ry122]

The way I interpret it is the following: There are three unknowns (the two tensions and the angle) and the constraint that neither tension must be over 10 kN.

The approach I would suggest would be to first assume that the tension T is the maximum value, 10 kN. And then solve for the other tension and the angle theta. If the other tension is less than 10 KN then that's it, that's the final answer to the problem.
If the other tension is above 10 kN, one has to start all over again by setting that other tension equal to 10kN and finding T and theta.

this is what I had in mind but i don't know how to solve for the other tension. Can you show me how to do that?

 

[Dick]

You have a direction (angle) and magnitude of each of the three forces acting on the pulley. Split each into x and y components. You'll need to use a trig function of the angles.

 

[Ry122]

This is what I came up with
sum of y components = 10cos0 +Fw + bcos30
sum of x components = bsin30 + 10sin0
I can't solve it because there's 3 unknowns

 

[Dick]

Ok, I'm guessing you are using b as the unknown tension and the symbol '0' to represent 'theta' (that's not a very good choice - just write e.g. cos(theta)). If that's the case then you are missing some signs. For example, one of the x forces should be positive and the other negative. They are pointing in opposite directions. Not all of the y forces are pointing in the same direction either. As for the other unknown 'Fw', the tensions on the two sides of the pulley are equal. So Fw isn't unknown, it's the same as T. Now there are two unknowns, b and theta.

 

[Ry122]

yes b = unknown tension
Assuming T and Fw are both negative forces then
-10cos(theta)-10+bcos30=0
bsin30-10sin(theta)=0
Is this correct?
The answer I got was 60 degrees.

 

[Dick]

Yes, the equations are correct. And, yes, I seem to get 60 degrees as well.

 

[Ry122]

No matter what value I have for T the answer is always 60 degrees. Should this matter?

 

[Dick]

No, you finally wind up with an equation for theta in which the T cancels out. This is a good thing. Is b>10kN? If so you will have to resolve the problem with b=10kN and then figure out what T is. If everything varies in proportion, this will save you a lot of time.

 

[Ry122]

Is there a way to solve for b without having to make theta the subject of the equation.

 

[Dick]

Is there a way to solve for b without having to make theta the subject of the equation.

Solve one equation for sin(theta) and the other for cos(theta). Square and add them. sin(theta)^2+cos(theta)^2=1. Poof, no more theta.

 

[Ry122]

-10cos(theta)-10+bcos30=0
bsin30-10sin(theta)=0

Im beginning to doubt that these equations are correct.
I think that either the -10 or the -10cos(theta) shouldn't be in there because they are the same force. Pulleys just change the direction of the force.
Does anyone agree with me?

 

[Dick]

Does anyone agree with me?

I don't. Those two terms are vertical components of two 10kN forces pointing in different directions. You were almost done with this problem. What's bothering you?