Tension of string joining two crossed sticks

1. Mar 8, 2015

Seaborgium

1. The problem statement, all variables and given/known data

"A cylinder of mass m and radius R is lodged between crossed sticks that make an angle θ with eachother. The crossed sticks, each of negligible mass, are connected at the point C, with AC=BC=2R and CD=CE=3R. Determine the tension in the string at DE. Assume the floor is smooth."

Associated diagram:

2. Relevant equations
Not sure if there are any, its a statics problem with no given data so it seems to be a puzzle of arranging force vectors.

3. The attempt at a solution

Establishing that the system is symmetrical, and that the string is light and inextensible and that tension is constant along it, I resolved to work with only one half of the system. Taking the contact point at the point on a stick where the cylinder meets it, the downward force would be mg (I would think) from which the normal contact force on the cylinder could be worked out in terms of mg and θ. From here, as the system is in equilibrium, the sum of forces along the horizontal plane would be zero, hence the horizontal component of the normal contact force would be equal to the tension in the string, thus completing the question.

I'm not confident that my understanding of the system is complete, and I'm not sure how to determine the normal contact force. Have I described all the forces in play correctly?

Thanks
Seaborg

2. Mar 8, 2015

PeroK

You can use symmetry to conclude that cefrtain forces must be equal. But I wouldn't then ignore half the forces. So, keep all forces in. Do a free-body diagram and write down all the forces involv ed and what you know about them and their relationship to each other.

Last edited: Mar 8, 2015
3. Mar 8, 2015

Seaborgium

I drew the diagram and noted as many forces as I could identify on the diagram - tensions, normals and gravity. I then drew a vector triangle out with the three vectors on, as Tension and Gravity are orthogonal to eachother in this case, giving me a triangle with the normal contact force as the hypotenuse. This is where everything fell apart as I defined the normal contact force as m.g.sin(θ) and the tension ended up with imaginary constant within it. Is vector triangle the wrong approach? Or have I missed some force out?

4. Mar 8, 2015

PeroK

As a side issue, where did you find this problem? I don't think the ball will fit! You've drawn it as best you can, but if you draw the ball to scale, the top of the ball will always be above the string. It would be okay if CD = 4R.

You've got two forces of mg instead of two forces of mg/2. Which means the answer you get should be for a ball of mass 2m.

You've jumped to the conclusion that $R = mgsin(\theta)$ You need to rethink that.

There are also forces on the ground.

You're missing something fundamental.

5. Mar 8, 2015

Seaborgium

I haven't drawn the thingy to scale - the top of the cylinder is above the string.

Fixed the error with mg/2. I thought it looked fishy when noting it on. Also, realised its mgcos(Φ) (for Φ = θ/2) and not sin.

The ground forces completely elude me though. There isn't any friction involved, as the ground is smooth. Guessing the normal contact force from the ground is what is missing?

6. Mar 8, 2015

PeroK

It's not $mgcos(\theta)$ either.

Yes, you're missing the normal forces on the ground.

What you're missing is that point C is a pivot. If point C was welded, you wouldn't need the string.

7. Mar 8, 2015

Seaborgium

I've got it! I need to resolve moments around the pivot for both sticks, which should be equal to zero as neither is rotating?

Makes a huge amount more sense now. Still no idea where to go on defining the normal contact force between the cylinder and the sticks, though.

8. Mar 8, 2015

PeroK

What's the vertical component of the normal contact force?

9. Mar 8, 2015

Seaborgium

Thinking it's mg/2. Would the contact force be (mg/2)cos(θ/2)?

10. Mar 8, 2015

PeroK

No, that's not it.

Which is larger? mg/2 or the contact force?

11. Mar 8, 2015

Seaborgium

mg/2 would be larger, seeing as the contact force is at an angle to the vertical that the mg/2 vector points along?

Sorry if I'm missing obvious stuff, I'm still a bit new to Physics.

12. Mar 8, 2015

PeroK

So, what's keeping the ball up?

13. Mar 8, 2015

Seaborgium

They're equal and opposite? Disappointed it took me that long to realise that.

So the normal contact force is equal to its horizontal and vertical component vectors added together - neither of which I feel like I understand how to find, even having done simpler static systems before. It's component perpendicular to a stick would be mg/2 cos(theta/2), but it would also have a component parallel to the stick - equal to the perpendicular component of the other half of the system?

14. Mar 8, 2015

haruspex

There are three forces acting on the cylinder. Your problem is that you are writing the statics equation as though the forces at a specific point of contact have to balance. They don't - it's the combination of all three forces that must balance. If you resolve normally to the contact with one stick then you must include a somewhat complicated contribution from the normal force at the other. If you think you've been resolving vertically, you haven't been getting the equation right; the mg will not get multiplied by the cos or sin of anything.

By the way, I believe there is flaw in the problem statement. It ought to specify that there is no friction between cylinder and sticks (or the problem becomes indeterminate).

Last edited: Mar 8, 2015
15. Mar 8, 2015

PeroK

You'll need to take care with your trigonometry on this one.

The contact force is normal to the stick. It has vertical and horizontal components. The vertical component must be mg/2.

16. Mar 8, 2015

Seaborgium

Oh Jesus, I see now. I've never seen a problem of this nature, so I'd assumed it would be like the others, without realising the cylinder wasn't simply experiencing one or two forces. The three forces acting upon it would be gravity, then a normal contact force from each of the sticks? Now I get why the vertical component of each normal contact force is mg/2, and the horizontal components must be equal and opposite - canceling each other out. Assuming this is the case and I'm not over-simplifying it, the normal contact force vertical component from the ground would be mg/2 (pointing up), so resolving the moment at C with respect to Tension in the rope and NCF from the ground would result in zero and define Tension in terms of trig, m and g?

17. Mar 8, 2015

haruspex

Yes, that should work.

18. Mar 8, 2015

Seaborgium

I ended up with T = mg/3 tan(θ/2) , which seems to make sense. Thanks to both of you for all the help, I feel like I understand this topic much better now!

19. Mar 8, 2015

PeroK

All three forces create a torque about C. I think you forgot the contact force.

20. Mar 8, 2015

Seaborgium

Oops, I went off the assumption that the horizontal components of the contact forces on the cylinder would cancel each other out and neglected to include them in the torque balance. This time I got T = mg/2 tan(θ/2). Does this look more correct?