How Do You Calculate the Tensions in a Two-Rope System with a Suspended Mass?

[PierceJ]

Homework Statement

A mass of 36.8kg hangs from a system of ropes
Both angles Θ = 23.7 deg

http://i.minus.com/i7ZTPchjaji7q.PNG

Find the magnitude of the tensions T₁ and T₂

Homework Equations

T₁sin(Θ)+T₂sin(Θ)=mg

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.

 

[SteamKing]

Homework Statement

A mass of 36.8kg hangs from a system of ropes
Both angles Θ = 23.7 deg

http://i.minus.com/i7ZTPchjaji7q.PNG

Find the magnitude of the tensions T₁ and T₂

Homework Equations

T₁sin(Θ)+T₂sin(Θ)=mg

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.

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

 

[PierceJ]

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

 

[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. $$\Sigma \vec{F}_{any direction} = m \vec{a}_{any direction}$$.

 

[PierceJ]

So:
T₁sinΘ = mg?

 

[Ashu2912]

So:
T₁sinΘ = mg?

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.

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. $$\Sigma \vec{F}_{any direction} = m \vec{a}_{any direction}$$.

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

 

[PierceJ]

T₁cosΘ+T₂cosΘ = 0
T₁sinΘ+T₂sinΘ = mg

Is this correct?

 

[PeroK]

Homework Statement

A mass of 36.8kg hangs from a system of ropes
Both angles Θ = 23.7 deg

http://i.minus.com/i7ZTPchjaji7q.PNG

Find the magnitude of the tensions T₁ and T₂

Homework Equations

T₁sin(Θ)+T₂sin(Θ)=mg

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.

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

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

 

[Ashu2912]

T₁cosΘ+T₂cosΘ = 0
T₁sinΘ+T₂sinΘ = mg

Is this correct?

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?

 

[PierceJ]

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?

Ah, so its T1cosΘ-T2cosΘ = 0?

Here's the key question: why must T₁=T₂?

Because the system is in equilibrium?

 

[PeroK]

Ah, so its T1cosΘ-T2cosΘ = 0?Because the system is in equilibrium?

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.

 

[Ashu2912]

Ah, so its T1cosΘ-T2cosΘ = 0?Because the system is in equilibrium?

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 ?

 

[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.

 

[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?

 

[PeroK]

Start with this equation:

T1cosΘ-T2cosΘ = 0

And show that $T_1 = T_2$

 

[PierceJ]

Okay so, you just move T₂ over.

 

[PeroK]

Okay so, you just move T₂ over.

What I would normally do now is:

Let $T = T_1 = T_2$

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

 

[PierceJ]

Alright.

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

T = 448.6N

 

[PeroK]

Alright.

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

T = 448.6N

That's correct.

 

[PierceJ]

Thank you for the help