Equilibrium of a Collar and Pulley System with Spring - Free Body Diagram

[Alison A.]

Homework Statement

A collar with a pulley slides on a frictionless vertical bar GH. A string A B C D is wrapped around, where portion AB of the string is horizontal. A spring with 2.5 lb/in. stiffness is placed between the collar and point H. The spring has 7 in. unstreched length and 5 in. final length. The collar and pulley have a combined weight of 3.1 lb, and the weights of all other components are negligible. Determine the value of P required for equilibrium, and the magnitude of the reaction between the collar bar GH.

Picture:
http://imgur.com/nHj2rrB

Hint given:
Before taking any other action, draw a FBD of the collar/pulley. This requires a closed external surface passing though the string at two locations as well as between the contact point between the spring and collar/pulley, and internal surface passing through the interface with the smooth bar. Pay close attention to the direction of the force exerted on the collar/pulley by the spring given the information in the problem statement.

Homework Equations

ΣF=0
ΣF_xi+ΣF_yj=0,
ΣF_x=0, and ΣF_y=0.
Spring Law: F_s=kδ=k(L-L₀)

The Attempt at a Solution

Here's my (attempted) FBD: http://imgur.com/YdaoEEW

I know it's probably incorrect, we never really did an example in class that involved springs. Also I wasn't sure what should have been in the middle B or C? B and C? Only B? Only C? Two body diagrams?

I know I need to add up all the x and y components then for P figure out the force it takes to make the x components equal 0. Then from there finding the magnitude of HG is just a matter of adding together their components and taking the magnitude.

I seem to have a really hard time figuring out the components, but when I find them I can pretty easily finish the problem.

Thank you so much for any help!

 

[billy_joule]

The Attempt at a Solution

Here's my (attempted) FBD: http://imgur.com/YdaoEEW

I know it's probably incorrect, we never really did an example in class that involved springs. Also I wasn't sure what should have been in the middle B or C? B and C? Only B? Only C? Two body diagrams?

I know I need to add up all the x and y components then for P figure out the force it takes to make the x components equal 0. Then from there finding the magnitude of HG is just a matter of adding together their components and taking the magnitude.

Having B&C in the middle as you've done is fine. You don't know the radius of the pulley anyway, not that it matters in this case.

Your diagram is not quite right. Which direction does the spring force act on the collar? Your diagram shows it pulling up on the collar (assuming H is the springs force).
It's also missing the reaction force from the bar HG.

You need to balance the y forces to find P, not the x. The reaction force of the bar HG will balance the x forces for whatever P ends up being.

 

[Alison A.]

Having B&C in the middle as you've done is fine. You don't know the radius of the pulley anyway, not that it matters in this case.

Your diagram is not quite right. Which direction does the spring force act on the collar? Your diagram shows it pulling up on the collar (assuming H is the springs force).
It's also missing the reaction force from the bar HG.

You need to balance the y forces to find P, not the x. The reaction force of the bar HG will balance the x forces for whatever P ends up being.

Does this look a little better?
http://imgur.com/dNw09kG

I'm not sure what you meant from the reaction force of bar HG

 

[billy_joule]

That's right.
You can go ahead and balance the Y forces now.

 

[Alison A.]

That's right.
You can go ahead and balance the Y forces now.

Alright so I don't get why the force of the spring would be negative
F_HC=(2.5)lb/in(5-7) which equals -5

 

[Alison A.]

W = 0 i + -3.1 j
T_CD = T_CD(4/5) i+T_CD(3/5) j
T_AB = ? i + 0 j
T_HC = 0 i + (2.5)(5-7) = -5 j

These don't seem right :/

 

[billy_joule]

That is almost correct.
Double check the x and y components of T_CD.
What can you say about tension in a string? That is, what relationship do T_CD and T_AB (and also P) have?
F_spring would be a better name than T_HC, T is usually reserved for tensions.

Alright so I don't get why the force of the spring would be negative
F_HC=(2.5)lb/in(5-7) which equals -5

It's just how Hookes law is defined. In this case it just so happens that for your choice of axes the negative spring force is correct. if the spring was compressed under, rather than above the collar the force would be positive. It's up to you to recognise what sign is appropriate.

From wiki: https://en.wikipedia.org/wiki/Hooke's_law

Consider a simple helical spring that has one end attached to some fixed object, while the free end is being pulled by a force whose magnitude is [PLAIN]https://upload.wikimedia.org/math/8/0/0/800618943025315f869e4e1f09471012.png. Suppose that the spring has reached a state of equilibrium, where its length is not changing anymore. Let

be the amount by which the free end of the spring was displaced from its "relaxed" position (when it is not being stretched). Hooke's law states that

or, equivalently,

where

is a positive real number, characteristic of the spring. Moreover, the same formula holds when the spring is compressed, with

and

both negative in that case. According to this formula, the graph of the applied force

as a function of the displacement

will be a straight line passing through the origin, whose slope is [PLAIN]https://upload.wikimedia.org/math/8/c/e/8ce4b16b22b58894aa86c421e8759df3.png.

 

[Alison A.]

That is almost correct.
Double check the x and y components of T_CD.

I get really confused on this whole triangle thing, could you help me out here?

What can you say about tension in a string? That is, what relationship do T_CD and T_AB (and also P) have?

They are the only vectors that have x-components?

F_spring would be a better name than T_HC, T is usually reserved for tensions.
It's just how Hookes law is defined. In this case it just so happens that for your choice of axes the negative spring force is correct. if the spring was compressed under, rather than above the collar the force would be positive. It's up to you to recognise what sign is appropriate.

From wiki: https://en.wikipedia.org/wiki/Hooke's_law

Alright this makes sense.

 

[billy_joule]

I get really confused on this whole triangle thing, could you help me out here?

T_CD is closer to vertical than horizontal, so should the magnitude of the x or the y component be greater?

(hint: swap the 3 and the 4 in your expression: TCD = TCD(4/5) i+TCD(3/5) j )

They are the only vectors that have x-components?

Nope, the reaction force between the bar GH and the collar also has (only) an x component.
Tension is equal throughout the string so T_CD = T_AB =P

https://en.wikipedia.org/wiki/Tension_(physics)

Tension in a string is a non-negative scalar. Zero tension is slack. A string or rope is often idealized as one dimension, having length but being massless with zero cross section. If there are no bends in the string (as occur with vibrations or pulleys), then tension is a constant along the string, equal to the magnitude of the forces applied by the ends of the string. By Newton's Third Law, these are the same forces exerted on the ends of the string by the objects to which the ends are attached. If the string curves around one or more pulleys, it will still have constant tension along its length in the idealized situation that the pulleys are massless and frictionless

 

[Alison A.]

T_CD is closer to vertical than horizontal, so should the magnitude of the x or the y component be greater?

(hint: swap the 3 and the 4 in your expression: TCD = TCD(4/5) i+TCD(3/5) j )

T_CD= T_CD (3/5) i + T_CD (4/5) j

Nope, the reaction force between the bar GH and the collar also has (only) an x component.
Tension is equal throughout the string so T_CD = T_AB =P

So if I find T_CD this equals P? How does this play in with the finding the magnitude of the reaction between collar and bar GH?

 

[billy_joule]

T_CD= T_CD (3/5) i + T_CD (4/5) j

Correct.

So if I find T_CD this equals P? How does this play in with the finding the magnitude of the reaction between collar and bar GH?

Yes.
In your FBD you have three forces in the x direction, they must sum to zero as the collar isn't accelerating. Once you have T_CD you can resolve the two known forces (T_{CD, X} and T_AB ) in the x direction to find the one unknown: F_{GH reaction}.

 

[Alison A.]

Correct.Yes.
In your FBD you have three forces in the x direction, they must sum to zero as the collar isn't accelerating. Once you have T_CD you can resolve the two known forces (T_{CD, X} and T_AB ) in the x direction to find the one unknown: F_{GH reaction}.

So to go about finding T_CD... Should I equal all the y-components together then solve for T_CD? Then plug that back into get the components? Then take the magnitude?

 

[billy_joule]

So to go about finding T_CD... Should I equal all the y-components together then solve for T_CD? Then plug that back into get the components? Then take the magnitude?

Yes. That's the general approach. Good luck.

 

[Alison A.]

Yes. That's the general approach. Good luck.

I'm going to assume T_spring is negative because it doesn't make sense any other way.
ΣF_Y=-3.1 + T_CD (4/5) + 0 - 5
This results in T_CD being 10.125...

Plugging into (3/5)(10.125)i +(4/5)(10.125)j = (6.075, 8.1)
The magnitude of this is 10.13lb... which is right

ΣF_X= 0 + 6.075 + x + 0
x = -6.075... but the x coordinate of T_AB is not in the negative direction

This is so frustrating

 

[billy_joule]

I'm going to assume T_spring is negative because it doesn't make sense any other way.
ΣF_Y=-3.1 + T_CD (4/5) + 0 - 5
This results in T_CD being 10.125...

Plugging into (3/5)(10.125)i +(4/5)(10.125)j = (10.125, 8.1)
The magnitude of this is 11.81lb... which isn't right

(3/5)(10.125)i ≠ 10.125i

But you had already found the magnitude of T_CD , it's 10.125 lb so that step was unnecessary.
you will need to find (3/5)(10.125)i though, to find the shaft reaction force.

 

[Alison A.]

(3/5)(10.125)i ≠ 10.125i

But you had already found the magnitude of T_CD , it's 10.125 lb so that step was unnecessary.
you will need to find (3/5)(10.125)i though, to find the shaft reaction force.

I editted my post, my bad, apparently I can't mutiply

 

[Alison A.]

Plugging into (3/5)(10.125)i +(4/5)(10.125)j = (6.075, 8.1)
The magnitude of this is 10.13lb... which is right

ΣFX= 0 + 6.075 + x + 0
x = -6.075... but the x coordinate of TAB is not in the negative direction

This is so frustrating

I guess what I'm not getting is exactly what the GH reaction is?

 

[billy_joule]

I guess what I'm not getting is exactly what the GH reaction is?

T_{CD, X} and T _AB are both pulling the collar to the right. But we know the collar doesn't move to the right because we are told it is in equilibrium.
So what is stopping the collar moving to the right?

This is a described by Newtons third law 'For every action there is an equal and opposite reaction'.
For whatever T_{CD, X} and T _AB are the bar will applies and equal and opposite force to the collar.

Just like when you push against a brick wall with your hand, no matter how hard you push, the wall won't move, it just pushes back at you with an equal and opposite force.

 

[Alison A.]

T_{CD, X} and T _AB are both pulling the collar to the right. But we know the collar doesn't move to the right because we are told it is in equilibrium.
So what is stopping the collar moving to the right?

This is a described by Newtons third law 'For every action there is an equal and opposite reaction'.
For whatever T_{CD, X} and T _AB are the bar will applies and equal and opposite force to the collar.

Just like when you push against a brick wall with your hand, no matter how hard you push, the wall won't move, it just pushes back at you with an equal and opposite force.

Didn't I just find T_{CD, X}? 6.075?

 

[billy_joule]

Didn't I just find T_{CD, X}? 6.075?

That's right.
So use your FBD and the fact that ∑F_x = 0 to find F_{GH reaction}

 

[Alison A.]

That's right.
So use your FBD and the fact that ∑F_x = 0 to find F_{GH reaction}

This is what I was having trouble with before... ∑F_x = 0 + 6.075 + T_{AB, X} + 0...
So T_{AB, X} = -6.075 but that doesn't make sense

 

[billy_joule]

This is what I was having trouble with before... ∑F_x = 0 + 6.075 + T_{AB, X} + 0...
So T_{AB, X} = -6.075 but that doesn't make sense

You already know the value of T_AB from post #9
Your, second (and correct) FBD has three forces in the x direction but your sum of x forces quoted has only two?

 

[Alison A.]

You already know the value of T_AB from post #9
Your, second (and correct) FBD has three forces in the x direction but your sum of x forces quoted has only two?

Since T_AB = P then it must be 10.13lb. I'm assuming the "reaction" is also the same the normal force of T_AB?

So ΣF_x = 6.075 + 10.13 - x (x being the x component of the normal vector which is what we are trying to find I think)

 

[billy_joule]

Since T_AB = P then it must be 10.13lb. I'm assuming the "reaction" is also the same the normal force of T_AB?

So ΣF_x = 6.075 + 10.13 - x (x being the x component of the normal vector which is what we are trying to find I think)

Yes, a normal force is a type of reaction force

https://en.wikipedia.org/wiki/Normal_force

F _{GH bar reaction} has only an x component as your second FBD shows and the problem statement implied when it stated the bar was frictionless.

 

[Alison A.]

Yes, a normal force is a type of reaction force

https://en.wikipedia.org/wiki/Normal_force

F _{GH bar reaction} has only an x component as your second FBD shows and the problem statement implied when it stated the bar was frictionless.

So all I had to do was add them together... ugh, so easy.

Thank you so much for helping me and especially being patient with me!

 

[billy_joule]

So all I had to do was add them together... ugh, so easy.

Yep, sometimes it's hard to see the forest for the trees.

Thank you so much for helping me and especially being patient with me!

You're welcome. Good work, your perseverance will get you far