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M1 tension on string question watch

1. Two small rings, A and B, each of mass 2m, are threaded on a rough horizontal pole. The
coefficient of friction between each ring and the pole is µ. The rings are attached to the ends of
a light inextensible string. A smooth ring C, of mass 3m, is threaded on the string and hangs in
equilibrium below the pole. The rings A and B are in limiting equilibrium on the pole, with
BAC = ABC = theta, where tan theta = 3/4

(a) Show that the tension in the string is 5/2 mg. done.

(b) Find the value of µ.

There has been a thread on this before but it was a long time ago. Basically I don't understand why tension is acting upward when you consider C. However when considering particles A and B the tension acts downward same direction as where the weight acts. Thanks.
2. (Original post by coconut64)
Two small rings, A and B, each of mass 2m, are threaded on a rough horizontal pole. The
coefficient of friction between each ring and the pole is µ. The rings are attached to the ends of
a light inextensible string. A smooth ring C, of mass 3m, is threaded on the string and hangs in
equilibrium below the pole. The rings A and B are in limiting equilibrium on the pole, with
BAC = ABC = theta, where tan theta = 3/4

(a) Show that the tension in the string is 5/2 mg. done.

(b) Find the value of µ.

Is tension acting upwards for C because the force of
There has been a thread on this before but it was a long time ago. Basically I don't understand why tension is acting upward when you consider C. However when considering particles A and B the tension acts downward same direction as where the weight acts. Thanks.
didn't finish your sentence eh? xD

I'm not too sure maybe because if there was no tension opposing the weight from particles A and B the string would just snap since there's no opposing force?

Maybe Zacken can provide and answer
3. (Original post by coconut64)
Two small rings, A and B, each of mass 2m, are threaded on a rough horizontal pole. The
coefficient of friction between each ring and the pole is µ. The rings are attached to the ends of
a light inextensible string. A smooth ring C, of mass 3m, is threaded on the string and hangs in
equilibrium below the pole. The rings A and B are in limiting equilibrium on the pole, with
BAC = ABC = theta, where tan theta = 3/4

(a) Show that the tension in the string is 5/2 mg. done.

(b) Find the value of µ.

There has been a thread on this before but it was a long time ago. Basically I don't understand why tension is acting upward when you consider C. However when considering particles A and B the tension acts downward same direction as where the weight acts. Thanks.
What do you mean by upwards and downwards here?

Do you mean vertically up and down, or along the string, in a general upwards/downwards direction with a horizontal component.

I.e. What's the actual issue for you? The fact that it's up at the centre and down at the ends, or the fact that its "vertical"?
4. (Original post by coconut64)
x
hey! coconut

i just foind this website which explains this really well, it's a few comments down and he has quite a few diagrams to explain where the tension goes
http://physics.stackexchange.com/que...-tension-force
5. (Original post by ghostwalker)
What do you mean by upwards and downwards here?

Do you mean vertically up and down, or along the string, in a general upwards/downwards direction with a horizontal component.

I.e. What's the actual issue for you? The fact that it's up at the centre and down at the ends, or the fact that its "vertical"?
Hi, the problem for me is that when considering particle A or B, I thought that tension would be acting up since the tension is acting upward for c. However, tension acts downward for AB.
6. (Original post by coconut64)
Two small rings, A and B, each of mass 2m, are threaded on a rough horizontal pole. The
coefficient of friction between each ring and the pole is µ. The rings are attached to the ends of
a light inextensible string. A smooth ring C, of mass 3m, is threaded on the string and hangs in
equilibrium below the pole. The rings A and B are in limiting equilibrium on the pole, with
BAC = ABC = theta, where tan theta = 3/4

(a) Show that the tension in the string is 5/2 mg. done.

(b) Find the value of µ.

There has been a thread on this before but it was a long time ago. Basically I don't understand why tension is acting upward when you consider C. However when considering particles A and B the tension acts downward same direction as where the weight acts. Thanks.
I got µ=4/7. Is that correct?
7. (Original post by coconut64)
Hi, the problem for me is that when considering particle A or B, I thought that tension would be acting up since the tension is acting upward for c. However, tension acts downward for AB.
OK, so your problem is really about how tension in a string acts.

It can be summed up in "tension acts towards the body of the string".

So, at the ends, it acts towards the middle.

At any other point on the string, the tension acts towards the bodies of the two pieces of the string either side of the point.

Hence at A or B it acts downwards in general, although it's actually acting along the line of the string, downwards at an angle.

At C, the tension acts upwards at an angle. One lot towards A, and one lot towards B. And because these two tensions are equal in magnitude, and at the same angle to the horizontal, but opposite sides, then their horizontal components cancel out and you're left with a net force acting vertically upwards.
8. (Original post by Steelmeat)
hey! coconut

i just foind this website which explains this really well, it's a few comments down and he has quite a few diagrams to explain where the tension goes
http://physics.stackexchange.com/que...-tension-force
Hi I think I sort of got the idea now thanks
9. is it unusual that answered the question correctly without even using the ring at C at all?

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