Another vector problem -- A ball suspended from 3 wires

[max1995]

Question

A ball with a mass 4 kg is suspended from 3 light in-extensible wires (diagram given)
Given that the ball is in equilibrium, calculate the tension in each of the wires.

My attempt

T₁ is equal to the weight of the ball which is 4g
so T1 = 4g

From there I don't know what to do but for T₃ is the vertical component equal to 4g and to find T₃ you can do 4gcos(70)= T₃?

is the above for T₃ right? If not how do I do it?

Not sure how to do T₂ could someone please explain?

Thanks

 

[Vagn]

Work with the horizontal components first. We know the horizontal components of T₂ and T₃ must be equal in magnitude. You can then relate T₂ and T₃.

 

[Simon Bridge]

You know the angles the tension forces make to each other, and you know one magnitude.
You know how to add vectors head-to-tail ... and you know what shape these vectors have to make when you do that.
You know some rules for finding lengths of sides when you know the angles.
(Or split everything into horizontal and vertical components...)

 

[max1995]

Work with the horizontal components first. We know the horizontal components of T₂ and T₃ must be equal in magnitude. You can then relate T₂ and T₃.

Thank you for the help :)

so horizontal of T₂=T₂cos60
horizontal of T₃= T₃sin70
so T₂cos60=T₃sin70 (is this right?)

 

[jtbell]

OK so far. Now, can you come up with a similar equation using the vertical components of T₁, T₂ and T₃?

 

[max1995]

OK so far. Now, can you come up with a similar equation using the vertical components of T₁, T₂ and T₃?

T₃cos(70)= vertical of T₃
T₂sin(60)= vertical of T₂
T₁=4g

Does this mean that for the verticals of the tensions 4g=T₃cos(70) + T₂sin(60)?

 

[jtbell]

4g=T3cos(70) + T2sin(60)?

Not quite. T₂ is upwards. T₁ and T₃ are downwards.

 

[max1995]

Not quite. T₂ is upwards. T₁ and T₃ are downwards.

Okay so the verticals of T₁ and T₃ must equal the vertical of T₂?

 

[jtbell]

Correct.

With horizontal components, the total to the left equals the total to the right.
With vertical components, the total upwards equals the total downwards.

(for a system in equlibrium, like this one.)

 

[max1995]

Correct.

With horizontal components, the total to the left equals the total to the right.
With vertical components, the total upwards equals the total downwards.

(for a system in equlibrium, like this one.)

I don't understand what to do after that because substituting in the equations means the tensions cancel out of the equation?

 

[jtbell]

You have two equations in two unknowns, T₂ and T₃. They should not cancel out when you solve the equations together. If they do, you've made a mistake in your algebra. Can you show your work?

 

[max1995]

You have two equations in two unknowns, T₂ and T₃. They should not cancel out when you solve the equations together. If they do, you've made a mistake in your algebra. Can you show your work?

Starting equations:
T₂cos(60)=T₃sin(70)
T₂(sin60) = T₃cos(70) + 4g

Then do I rearrange one of them to make one of the T's the subject and sub it into the other equation?Thank you for all the help btw :)

 

[jtbell]

Then do I rearrange one of them to make one of the T's the subject and sub it into the other equation?

Exactly. :D

 

[max1995]

Exactly. :D

Would you be able to give me an example of how to rearrange and substitute these equations to find one of the tensions? as I am having trouble doing this

 

[jtbell]

Show us what you've tried and where you get "stuck" and I or someone else can probably help you get "unstuck."

 

[max1995]

Show us what you've tried and where you get "stuck" and I or someone else can probably help you get "unstuck."

Starting equations:
T₂cos(60)=T₃sin(70) (Equation 1)
T₂(sin60) = T₃cos(70) + 4g (Equation 2)

Rearranging T₂(sin60) = T₃cos(70) + 4g to get T₂ = (T₃cos(70) + 4g)/sin(60) then substitue it into equation 1

this gives (T₃cos(70) + 4g)/sin(60)xcos(60) = T₃sin(70)

then divided both sides by cos(60) to give

(T₃cos(70) + 4g)/sin(60) = (T₃sin(70))/cos60

Times both sides by (sin60)

(T₃cos(70) + 4g = [(T₃sin(70))/cos60]x sin60

this gives in numerical form

0.3420201433T₃ + 39.24 = 1.627595363T₃

Then subtract 0.3420201433T₃ from both sides

39.24 = 1.28557522T₃

then divde both sides by 1.28557522

T₃ = 30.52N

for T₃ this is what I got, is it correct?

 

[jtbell]

I'm starting to look over your algebra, but you can check the final answers yourself. Use your value of T₃ to find T₂, then substitute both into your starting equations (at the top of your post) and see if they work out.

 

[jtbell]

It looks OK to me so far. I wasn't sure at first because in your first post I missed the mass of the ball being "4 kg" and assumed that by "4g" in your equations you meant 4 grams. So I got a different number for T₃ at first. I fixed that and now we agree.

 

[max1995]

It looks OK to me so far. I wasn't sure at first because in your first post I missed the mass of the ball being "4 kg" and assumed that by "4g" in your equations you meant 4 grams. So I got a different number for T₃ at first. I fixed that and now we agree.

yeah sorry i mean 4 times by gravity when saying 4g
Okay thank you for all the help :D

to find T2 is it alright just to sub in my value of T3 into one of my first equations to get its value?

Also is there a rep system on here so I can rep you? (new to this forum)

 

[jtbell]

o find T2 is it alright just to sub in my value of T3 into one of my first equations to get its value?

Right. Either equation should work. If you try both of them and get the same value for T₂, that will also serve as a check.

is there a rep system on here

Not the kind that I think you're thinking of. We have other ways of identifying the "good guys."

 

[Simon Bridge]

There's a "like" button below threads.

I like the geometric approach.

You have ##\vec T_1 + \vec T_2 +\vec T_3 = 0##
Draw those head-to-tail and you get a triangle.
Sketch out a scalene triangle but the ##T_1## side is vertical - try to get the directions approximately right.
Label the triangle like you would in maths class: the sides are a-b-c going clockwise around from ##T_1##
The angles opposite the sides are labelled with the capitol letters so A is opposite a etc.

So:
##a=T_1=mg,\; b=T_2,\; c=T_3## ... easy.
The angles have to be worked out carefully ...
c makes 70deg to the vertical, so the external angle between c and a is 70deg, so B=180-70=110
b makes 60deg to the horizontal, which is 150deg to the vertical, so the exterior angle between a and b is 150deg, so C=180-150=30deg
That means that A=180-110-30=40deg
Now you can use the sine rule to find ##T_2## and ##T_3##

You'd actually end up with the same equatons - only faster with less chance of making an algebraic slipup.