Calculating Weight of Collar in Equilibrium with Spring Force

[838]

Homework Statement

Collar A can slide on a frictionless vertical rod and is attached as shown to a spring. The constant of the spring is 4lb/in., and the spring is unstretched when h=12in. Knowing that the system is in equilibrium when h=16in., determine the weight of the collar.

Home-made diagram:

http://img182.imageshack.us/img182/9642/staticsuc5.th.png

Homework Equations

F=-kx
k=4lb/in
x=16in-12in=4in
F=-kx=-(4lb/in)*(4in)=16lbs

The Attempt at a Solution

With the pulley, the spring would still be carrying 100% of the collar's weight.

16lbs for the collar weight can't be the answer, can it?

Any help would be appreciated, it's been a while since I've taken a course in physics.

 

[alphysicist]

Hi 838,

Homework Statement

Collar A can slide on a frictionless vertical rod and is attached as shown to a spring. The constant of the spring is 4lb/in., and the spring is unstretched when h=12in. Knowing that the system is in equilibrium when h=16in., determine the weight of the collar.

Home-made diagram:

http://img182.imageshack.us/img182/9642/staticsuc5.th.png

Homework Equations

F=-kx
k=4lb/in
x=16in-12in=4in
F=-kx=-(4lb/in)*(4in)=16lbs

The Attempt at a Solution

With the pulley, the spring would still be carrying 100% of the collar's weight.

If the string were pulling vertically upwards, then the spring force would be equal to the collar's weight.

However, in this case there is another force besides the spring force and gravity that must be canceled for the collar to be in equilibrium. What is that force?

Do you now see how to relate the spring force and the weight?

 

[838]

The tension force from the rope?

Also, the pulley would cut the tension in half, so there would be 8lbs on each side, correct?

The angle that the pulley (B) makes with the collar (A) is 53.13 degrees, so would I break it into x and y components? Ah, no, that makes no sense.

I am thoroughly confused, this is probably an extremely simple problem that I'm just complicating.

 

[alphysicist]

The tension force from the rope?

Also, the pulley would cut the tension in half, so there would be 8lbs on each side, correct?

With the pulley massless and frictionless, the tension along the rope will be the same and equal to the spring force. So the question is how to relate the tension force to the weight.

The angle that the pulley (B) makes with the collar (A) is 53.13 degrees, so would I break it into x and y components?

That's the right idea. The tension is pulling on the collar at an angle. So it's pulling up and to the right. If you draw a free body diagram and apply Newton's law to the horizontal and vertical directions, you can see how the different forces cancel each other out.

In other words, once you have the vertical and horizontal components of the tension, what other force is the vertical component counteracting, and what other force is the horizontal component counteracting?

 

[838]

Ok, so I've split the force components. 6.4lbs in the y direction, and 4.8lbs in the x direction.

The vertical component is opposing gravity and the horizontal component is opposingthe bar that is holding the collar? This seems like it would make sense, since the system would be in equilibrium when all these forces cancel.

Would the collar weigh 6.4lbs?

 

[alphysicist]

Ok, so I've split the force components. 6.4lbs in the y direction, and 4.8lbs in the x direction.

The vertical component is opposing gravity and the horizontal component is opposingthe bar that is holding the collar? This seems like it would make sense, since the system would be in equilibrium when all these forces cancel.

Would the collar weigh 6.4lbs?

The tension is not cut in half by the pulley; it's magnitude is equal to the spring force (16 lbs). What then are the $x$ and $y$ components?

 

[838]

The tension is not cut in half by the pulley; it's magnitude is equal to the spring force (16 lbs). What then are the $x$ and $y$ components?

Ah, my mistake, I wrote 8lbs on my paper and neglected to erase.

x comp=16*sin(36.86)=9.6lbs
y comp=16*cos(36.86)=12.8lbs

So the collar weighs 12.8lbs.

 

[alphysicist]

Ah, my mistake, I wrote 8lbs on my paper and neglected to erase.

x comp=16*sin(36.86)=9.6lbs
y comp=16*cos(36.86)=12.8lbs

So the collar weighs 12.8lbs.

I was looking over how you got the force of 16lbs in the first place, and I think that's not quite right. The stretch in the spring is related to the change in the length of the hypotenuse.

However, the 4in increase is the increase in the length of the vertical leg. Using the Pythagorean theorem on the before and after triangle will let you find the change in length of the hypotenuse, and it's that change that will represent the stretch of the spring.

Once you find the new force (which will be somewhat less than 16lbs), I believe you can follow the same procedure as above and get the correct answer.

 

[838]

So, the hypotenuse BA when h=12 is 16.97, and when h=16 it is 20.

To find the spring force, F=-k(xo-xf) =-4lb/in(16.97in-20in)=12.12lbs
Now, to find the angle, which would be sin($$\theta$$)=12/20 so, $$\theta$$ = 36.869 degrees.

Now, splitting components, y=12.12*cos(36.869)=9.696lbs = vertical force.
x=12.12*sin(36.869)=7.32lbs = horizontal force.

So the weight of the collar should be 9.7lbs.


Thank you so much for your input. I'm sorry I'm a little slow when it comes to physics, like I said, it's been a while.