# Torque and tension question

1. Apr 20, 2015

### goonking

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

2. Relevant equations

3. The attempt at a solution
seems like Torque = r f sin theta should be used here but we don't know r.

Am I suppose to find the x and y components of the tension of the chain?

off topic : if the chain was moved so it would just be above the end of the rod, just like this : , would it put less tension on the chain?

2. Apr 20, 2015

### BvU

For variables you don't know, adopt a symbol and go to work. Good chance the thing will drop out towards the end.

3. Apr 20, 2015

### Orodruin

Staff Emeritus
You do not always need to know all of the quantities. Why don't you start by simply assuming that the length of the rod is R and do things analytically from there. A good practice is to never try to insert numbers until you have arrived at your final expression.

4. Apr 20, 2015

### Delta²

No the chain is in the proper angle to help you calculate the torque of the tension of chain easily $T_1=31.2r$ where r is the length of rod. This torque will be countered by the torque of the weight. Can you find what is the angle $\phi_1$ that the weight makes with the line of rod and then find the torque of weight? You then can see that r is simplified when equating the two torques.

5. Apr 20, 2015

### BvU

Problem solving systematics urges you to list the relevant equations and the given/known data.
You do know them already and you do use them already, but still it's a good way to get a grasp on an exercise.

In this case you want $\Sigma\;\tau = 0$ and indeed need $\tau = \vec r\times\vec F = r F \sin\theta$ but this $\theta$ is not the $\theta$ in the drawing that comes with your exercise.

Oh, am I a slow typist. And I need to do some paid work too, so I leave you to the other good helpers...

6. Apr 20, 2015

### goonking

i'm not sure where you mean ϕ1. by weight, do you mean the right end of the rod?

7. Apr 20, 2015

### Delta²

No i mean the force of weight due to gravity... Can you draw a diagram with all the forces acting on the rod?

8. Apr 20, 2015

### goonking

should be 78 degrees

9. Apr 20, 2015

### Delta²

Ok correct, now what is the torque of weight?

10. Apr 20, 2015

### goonking

r f sin 78

f = mg

?

11. Apr 20, 2015

### Delta²

Not exactly. The force of weight is considered to be applied at the c.o.m of the rod. Where is the c.o.m of rod?

12. Apr 20, 2015

### goonking

at the middle of the rod

13. Apr 20, 2015

### Delta²

ok so you have the weight acting on the middle of the rod, at an agle of 78, so what is its torque?

14. Apr 20, 2015

### goonking

r/2 x mg x sin 78

?

15. Apr 20, 2015

### Delta²

yes correct.

16. Apr 20, 2015

### goonking

do we need the torque of the chain? or is that given by the tension in the chain?

17. Apr 20, 2015

### Delta²

yes you need the torque of the chain. I already said in post #4 what it is. Equate the two torques and solve for m. What do you get?

18. Apr 20, 2015

### goonking

torque of chain = r 31.2

r 31.2N = r mg sin 78

m = 3.544kg

?

19. Apr 20, 2015

### Delta²

Didnt we say the torque of weight is (r/2)mgsin(78)

20. Apr 20, 2015

whoops, yes