Tension

[golden]

Could someone please explain Question 267 from the 2011 Checkpoints (or Question 270 from the 2010 Checkpoints) (same question)?

I've looked at the solutions but it doesn't explain completely what to do etc. besides use a formula.

The question goes:
Two poles with two strings off each pole are attached to a cage at the centre. The cage goes in a circular path of radius 8.0 m, and the strings make an angle of 30 degrees with the vertical (poles) at the highest point of the journey. In addition, the mass of the cage etc. is 250 kg and the speed of the cage at the bottom top of the path is 10 m/s (8 m/s at the top of the ride).
Calculate the tension (should be around 360N) at the top of the circle (for each string which should be the same).


2Tcos/(theta) + mg = Fnet
How was this derived?

*Main edits in red.

[xZero]

Lets look at the tension of 1 string only, it has a vertical and horizontal component. The vertical component is the one we need to worry about since the direction of motion is in the vertical direction. The vertical component of the tension is Tcos(theta) and since there are 2 strings, we double that, hence the 2Tcos(theta).

 At the bottom of the motion, the upward force is the vertical component of both strings and the downward force is the weight force (mg). Hence Fnet = 2Tcos(theta)-mg (note i used -mg because i defined my g as 9.8 instead of -9.

[schnappy]

Recognise that the net force is a centripetal force.
F_net = F_c = mvv/r = 3125 N

Look at the bottom of the motion, the net force acting on the cage must be F_c. Ignore F_c, and just list the forces on the cage:
Tension + Weight
(where tension is 'up', and the weight is 'down').
2Tcos(30) + mg = F_net
F_net = F_c
F_c = 2Tcos(30) + mg
3125 = 2Tcos(30) + mg

And substitute (g=-10, m=250):
3125 = 2Tcos(30) - 2500
T=3248 N.

And I've just realised as I went to post it asked for the tension at the top...

So if we look at the top, the net force is still a centripetal force, but it's made up a bit differently:
F_c = mg + 2Tcos(30)
(The velocity is still 10 m/s, F_c is still the same)
Because if you draw a diagram with all the forces acting on the cage, the tension and the weight are both going down. For simplicity I'll keep everything positive and say that positive is the down direction.
3125 = 2500 + 2Tcos(30)
T = 361 N

[golden]

Thank you both of you for your response.

One quick thing to go over:
How do we know if the tension is up or down at specific points besides from knowing that mg = 2500 and T = 3125 (hence T must be down)?

I believe that sometimes being more conceptual helps solve the problem in the first place as opposed to doing calculations (sometimes) to solve the problem.

Thanks!

[schnappy]

T != 3125, 3125 is the centripetal (net) force, not the tension.

Just remember that the centripetal force has to be made up of other things, albeit gravity, weight, tension, normal force.

So if we look at the top, the cage is accelerating towards the centre of the circle, as per the centripetal force. You just have to ask yourself how that centripetal force is made up... obviously tension, but the mass is also there, and at that point in the motion the weight force is in the exact same direction as the centripetal force. The centripetal is the net force (the result of all the acting forces), so we say that F_c = (Force by tension) + (weight)

If you want to know the direction of the tension force acting on the cage you can just use the equations and see if you get a positive or negative answer, but admittedly that can get confusing (Especially when g can be +-10 depending on what you're doing and how you've set things up). Other than that all I can really say is to just think about it really... at the bottom something is pulling it up, so the wires must be pulling it up. At the top it's coming down, it's weight is bringing it down, but the tension is as well. Tension can be less because the weight is taking over a bit.

[Bozo]

That question was a !@#$%

[schnappy]

It's better than obscure conceptual electronics questions.

[enpassant]

Could someone please explain Question 267 from the 2011 Checkpoints (or Question 270 from the 2010 Checkpoints) (same question)?

I've looked at the solutions but it doesn't explain completely what to do etc. besides use a formula.

The question goes:
Two poles with two strings off each pole are attached to a cage at the centre. The cage goes in a circular path of radius 8.0 m, and the strings make an angle of 30 degrees with the vertical (poles) at the highest point of the journey. In addition, the mass of the cage etc. is 250 kg and the speed of the cage at the bottom of the path is 10 m/s.
Calculate the tension (should be around 360N) at the top of the circle (for each string which should be the same).


2Tcos/(theta) + mg = Fnet
How was this derived?

What type of vertical circular motion is it? Your information did not indicate the type. With just the 2 strings holding the cage, can it be UNIFORM circular motion?

[golden]

Could someone please explain Question 267 from the 2011 Checkpoints (or Question 270 from the 2010 Checkpoints) (same question)?

I've looked at the solutions but it doesn't explain completely what to do etc. besides use a formula.

The question goes:
Two poles with two strings off each pole are attached to a cage at the centre. The cage goes in a circular path of radius 8.0 m, and the strings make an angle of 30 degrees with the vertical (poles) at the highest point of the journey. In addition, the mass of the cage etc. is 250 kg and the speed of the cage at the bottom of the path is 10 m/s.
Calculate the tension (should be around 360N) at the top of the circle (for each string which should be the same).


2Tcos/(theta) + mg = Fnet
How was this derived?

What type of vertical circular motion is it? Your information did not indicate the type. With just the 2 strings holding the cage, can it be UNIFORM circular motion?

Thanks for that, it's not uniform circular motion.

But if it was, would the answer change?

[enpassant]

If the speed is 10 at the bottom of the path, the cage cannot reach the top of the circular path according to conservation of energy (kinetic and gravitational). The strings do not contribute energy to the motion because the total force of the 2 strings on the cage is always normal to the path.

[schnappy]

It is uniform circular motion, that's the assumption you have to make. Any change in the magnitude of the acceleration would have to be given or work-outable, so you can find the velocity at the top of the circle. Or they could just give you the v at the top of the circle... None of this is given so just assume that the speed is constant hence it is uniform.

[golden]

It is uniform circular motion, that's the assumption you have to make. Any change in the magnitude of the acceleration would have to be given or work-outable, so you can find the velocity at the top of the circle. Or they could just give you the v at the top of the circle... None of this is given so just assume that the speed is constant hence it is uniform.

I like your thinking.

I should have put:
10 m/s at the top of the ride.
8 m/s at the bottom of the ride.

I've made that edit to the first post.

I didn't think that the speed and uniform circular motion would have been brought up .