# Homework Help: Mass hung from two ropes

1. Mar 4, 2015

### PierceJ

1. The problem statement, all variables and given/known data
A mass of 36.8kg hangs from a system of ropes
Both angles Θ = 23.7 deg

http://i.minus.com/i7ZTPchjaji7q.PNG [Broken]

Find the magnitude of the tensions T1 and T2

2. Relevant equations
T1sin(Θ)+T2sin(Θ)=mg

3. The attempt at a solution
I think you have to use the horizontal component of the tensions in this, but other than that, I am not really sure what to do here. Any help is appreciated.

Last edited by a moderator: May 7, 2017
2. Mar 4, 2015

### SteamKing

Staff Emeritus
What about the vertical components? Don't they come into play as well?

Last edited by a moderator: May 7, 2017
3. Mar 4, 2015

### PierceJ

Sorry, I meant to say that the horizontal bit is involved with the vertical bit somehow.

4. Mar 4, 2015

### Ashu2912

No, they both are independent equations. You have to resolve(split vectorially) the forces in the "direction along which the acceleration component is known", which here is all possible directions, as the object is in equilibrium. These directions may be mutually perpendicular but it is not always compulsory to do so, as in this case.
$$\Sigma \vec{F} = m \vec{a}$$ implies that we can take components in any directions i.e. $$\Sigma \vec{F}_{any direction} = m \vec{a}_{any direction}$$.

5. Mar 4, 2015

So:
T1sinΘ = mg?

6. Mar 4, 2015

### Ashu2912

What made you think that?

No, you have three forces T1 ,T2 and the weight mg. Resolve these in the vertical and horizontal directions and you will get two equations.

What I meant here was that
(1) The equations you get are independent
(2) It is not always required to resolve in two mutually perpendicular directions

7. Mar 4, 2015

### PierceJ

T1cosΘ+T2cosΘ = 0
T1sinΘ+T2sinΘ = mg

Is this correct?

8. Mar 4, 2015

### PeroK

To go back to your first post, you have almost solved this.

Here's the key question: why must $T_1 = T_2$?

Last edited by a moderator: May 7, 2017
9. Mar 4, 2015

### Ashu2912

The first equation is not. When you resolve T1 and T2 in the horizontal direction, you get T1 cosθ and T2cosθ, but in which directions?

10. Mar 4, 2015

### PierceJ

Ah, so its T1cosΘ-T2cosΘ = 0?

Because the system is in equilibrium?

11. Mar 4, 2015

### PeroK

Not just equilibium. Also, because the system is Symmetrical.

You've done the maths now with the horizontal forces to show that $T_1 = T_2$, but you should also try to "see" that $T_1 = T_2$ by symmetry. When you come to more complex problems, using symmetry in a system can be a very powerful tool/shortcut.

12. Mar 4, 2015

### Ashu2912

Correct, since the horizontal acceleration is zero (system is in equilibrium), the (vector) sum of forces resolved along the horizontal must be zero. You have T1cosθ towards left and T2cosθ towards right.

Now, if a body is in equilibrium, it's acceleration is zero implies the vector sum of the forces on it is zero(vector). Here, there are three forces in play. Thus the vector sum of T1, T2 and mg is zero. It just so happens that T1=T2 in this case. What if the angles were different, say θ1 and θ2 ?

13. Mar 4, 2015

### PeroK

There are two approaches to this problem. Neither is more correct than another, but see what you think:

1) You do some maths and find that, hey, $T_1 = T_2$.

2) You see that $T_1 = T_2$ by symmetry and look for the maths that will prove/confirm this.

14. Mar 4, 2015

### PierceJ

If the angles were different, they would still add up to be mg right?

As far as the answer goes for this question, what do I do here exactly?

15. Mar 4, 2015

### PeroK

T1cosΘ-T2cosΘ = 0

And show that $T_1 = T_2$

16. Mar 4, 2015

### PierceJ

Okay so, you just move T2 over.

17. Mar 4, 2015

### PeroK

What I would normally do now is:

Let $T = T_1 = T_2$

Then, go back to your first equation and solve for T.

18. Mar 4, 2015

### PierceJ

Alright.

T1sinΘ+T2sinΘ = mg
T(sinΘ+sinΘ) = mg
T = mg/2sinΘ

T = 448.6N

19. Mar 4, 2015

### PeroK

That's correct.

20. Mar 4, 2015

### PierceJ

Thank you for the help