Static equilibrium pulley problem

[virtuoso735]

Homework Statement

[PLAIN]http://img849.imageshack.us/img849/3716/staticeqpulley.jpg

Hopefully the image showed up. If not, here's the link: http://img707.imageshack.us/img707/3716/staticeqpulley.jpg"

Sorry for the low quality pictures. Hopefully it will do though. The question is: Relate the traction force F₃ to the weight mg and the angle theta.

Homework Equations

They give us that the tension T is uniform throughout the string (i.e. F₁, F₂ = T). F₃ acts along the horizontal.

The Attempt at a Solution

I resolved the diagonal forces into horizontal and vertical components. Since θ₁ and θ₂ are equal I just call them both θ. The system is in static equilibrium so the total force is zero. So I expand the equilibrium equations out as follows:

-F₁cosθ - F₂cosθ + F₃ = 0

and

-F₁sinθ + F₂sinθ = 0

Also, T = mg

Do I just sum the two equations above and set them equal to mg? That doesn't make sense since the sum is 0, and mg is certainly not 0. Or should I sum them and subtract mg from the sum?

Any help would be appreciated! Thanks.

 

[PhanthomJay]

Hi, virtuoso735, welcome to PF!

Homework Statement

The question is: Relate the traction force F₃ to the weight mg and the angle theta.

Homework Equations

They give us that the tension T is uniform throughout the string (i.e. F₁, F₂ = F₃).

F1 = F2, but F3 is not part of the string. F3 does not = F1 or F2.

F₃ acts along the horizontal.

yes.

The Attempt at a Solution

I resolved the diagonal forces into horizontal and vertical components. Since θ₁ and θ₂ are equal I just call them both θ. The system is in static equilibrium so the total force is zero. So I expand the equilibrium equations out as follows:

-F₁cosθ - F₂cosθ + F₃ = 0

correct.

and

-F₁sinθ + F₂sinθ = 0

yes, but since F1 = F2, this just tells you that 0 = 0, a well known fact. . But it also tells you that θ1 = θ2, something you assumed or realized earlier.

Also, T = mg

yes, and T = F₁ = F₂

Do I just sum the two equations above and set them equal to mg? That doesn't make sense since the sum is 0, and mg is certainly not 0. Or should I sum them and subtract mg from the sum?

Any help would be appreciated! Thanks.

Try it again, and see what you get.

 

[virtuoso735]

Is this right?

I replace the F₁ and F₂ with T. So I get -Tcosθ - Tcosθ + F₃ - mg = 0.

-2Tcosθ + F₃ = mg

Then I replace the T with mg, so -2mgcosθ + F₃ = mg

Is this the relationship?

 

[PhanthomJay]

Is this right?

I replace the F₁ and F₂ with T. So I get -Tcosθ - Tcosθ + F₃ - mg = 0.

-2Tcosθ + F₃ = mg

Then I replace the T with mg, so -2mgcosθ + F₃ = mg

Is this the relationship?

No. You are looking at forces in the x direction. You had the right equation originally, but now you have added an mg term. Substitute 'T' with 'mg'.

 

[virtuoso735]

Sorry, I'm not understanding what you mean. Which was the right equation I had originally? Should I not have added the two cosine terms together? So is -mgcosθ - mgcosθ + F3 - mg = 0 any good?

 

[ehild]

Your original equation for the horizontal force components was correct. "-F1cosθ - F2cosθ + F3 = 0"

You know that F1=F2=T and T=mg. And you need F3 in terms of mg and theta.

ehild

 

[virtuoso735]

Does that work out to F₃ = 2mgcosθ ?

Here is another similar problem, which I used the same process on:

[PLAIN]http://img708.imageshack.us/img708/5553/pulley2q.jpg

There are horizontal and vertical components to this problem as well.

For the horizontal component I get: -F1cosθ + F2 = 0.
For the vertical component I get: F1sinθ = 0 [Is this right? This component is positive since it is going up, right? Should it be equal to 0]

Summing them up gives: -F1cosθ + F2 + F1sinθ = 0.
Simplify to: -mgcosθ + F2 + mgsinθ = 0
mg(sinθ - cosθ) = F2

Does this make sense?

Also, how would the mass of the pulley affect the angle?

 

[PhanthomJay]

Does that work out to F₃ = 2mgcosθ ?

Yes!

Here is another similar problem, which I used the same process on:

[PLAIN]http://img708.imageshack.us/img708/5553/pulley2q.jpg

There are horizontal and vertical components to this problem as well.

You have given us the sketch, but not the question.

For the horizontal component I get: -F1cosθ + F2 = 0.

That should be sin, not cos, otherwise this is correct.

For the vertical component I get: F1sinθ = 0 [Is this right? This component is positive since it is going up, right? Should it be equal to 0]

It should not be equal to 0, and it's cos, not sin. You have F1 cos theta acting up, but what force acts down? Their sum is equal to 0.

Summing them up gives: -F1cosθ + F2 + F1sinθ = 0.
Simplify to: -mgcosθ + F2 + mgsinθ = 0
mg(sinθ - cosθ) = F2

Does this make sense?

No.

Also, how would the mass of the pulley affect the angle?

It would be less.

 

[virtuoso735]

Sorry, the question is: relate the traction force F2 to the weight mg and the angle theta.

I tried it again, hopefully this is closer.

Horizontal component: -F1sinθ + F2 = 0
Vertical component: F1cosθ - mg = 0

Substitute F1 and F2 with mg and sum them, which leaves:
-mgsinθ + F2 + mgcosθ - mg = 0

Solving for F2:
F2 = mg(sinθ - cosθ + 1)

Is this right?

 

[virtuoso735]

Sorry, the question is: relate the traction force F2 to the weight mg and the angle theta.

I tried it again, hopefully this is closer.

Horizontal component: -F1sinθ + F2 = 0
Vertical component: F1cosθ - mg = 0

Substitute F1 and F2 with mg and sum them, which leaves:
-mgsinθ + F2 + mgcosθ - mg = 0

Solving for F2:
F2 = mg(sinθ - cosθ + 1)

Is this right?

 

[PhanthomJay]

Sorry, the question is: relate the traction force F2 to the weight mg and the angle theta
I tried it again, hopefully this is closer.

Horizontal component: -F1sinθ + F2 = 0
Vertical component: F1cosθ - mg = 0

Substitute F1 and F2 with mg and sum them, which leaves:
-mgsinθ + F2 + mgcosθ - mg = 0

Solving for F2:
F2 = mg(sinθ - cosθ + 1)

Is this right?

No.
If the pulley is frictionless, the tension in the string wrapped around it will be equal...this should have been stated in the problem. So F2 is equal to mg, but F1 is not (it's a different string). You should start to get used to drawing free body diagrams.

the question is: relate the traction force F2 to the weight mg and the angle theta.

And the answer is...?