The Student Room Group

Mechanics - Tension in a string

Figure 2 shows a washing line suspended at either end by vertical rigid poles. A jacket of
mass 0.7 kg is suspended in equilibrium part of the way along the line. The sections of the
washing line on either side of the jacket make angles of 35° and 40° with the horizontal.


Right, so I thought the tension would be the same either side of the string. But apparently not.. can someone explain why? i mean, its the same string :/ i dont see why the string should have different tensions either side!
@Zacken
Reply 1
Original post by Someboady
Figure 2 shows a washing line suspended at either end by vertical rigid poles. A jacket of
mass 0.7 kg is suspended in equilibrium part of the way along the line. The sections of the
washing line on either side of the jacket make angles of 35° and 40° with the horizontal.


Right, so I thought the tension would be the same either side of the string. But apparently not.. can someone explain why? i mean, its the same string :/ i dont see why the string should have different tensions either side!
@Zacken


Because they are at different angles. Had they beena t the same angle, their tensions would be the same.
Reply 2
Original post by Zacken
Because they are at different angles. Had they beena t the same angle, their tensions would be the same.


Ah, thanks!
Learnt something new about mechanics about a month before my exam.. scary stuff :/
Reply 3
Original post by Someboady
Ah, thanks!
Learnt something new about mechanics about a month before my exam.. scary stuff :/


Generally, things are only the same if you have symmetry in the problem, which isn't quite so in this case because the angles ruin the symmetry. Rather a month before than a day after!
Reply 4
Original post by Zacken
Generally, things are only the same if you have symmetry in the problem, which isn't quite so in this case because the angles ruin the symmetry. Rather a month before than a day after!

One more small question...
If the situation changes but the slope stays the same... is the coefficient of friction the same? e.g.
A particle of mass 0.8 kg is held at rest on a rough plane
a) Calculate the co-efficient of friction
The particle is now held on the same rough plane
b) calculate something else

Can I assume the coefficient friciton is the same?
Reply 5
Original post by Someboady
One more small question...
If the situation changes but the slope stays the same... is the coefficient of friction the same? e.g.
A particle of mass 0.8 kg is held at rest on a rough plane
a) Calculate the co-efficient of friction
The particle is now held on the same rough plane
b) calculate something else

Can I assume the coefficient friciton is the same?


Yep, the coefficient of friction depends only on the material of the plane and you are clearly told that it is the same rough plane.
Reply 6
Original post by Zacken
Because they are at different angles. Had they beena t the same angle, their tensions would be the same.

If we use the normal M1 modelling assumptions then wouldn't the angles be the same?

Seems like quite a strange question unless I'm misiing something.
Reply 7
Original post by notnek
If we use the normal M1 modelling assumptions then wouldn't the angles be the same?

Seems like quite a strange question unless I'm misiing something.


Probably me being thick, but I don't quite understand what you're saying?
Reply 8
Original post by Zacken
Probably me being thick, but I don't quite understand what you're saying?



Could a mass really hand like this (on a single washing line) in equilibrium with angle a different to angle b?
Reply 9
Original post by notnek


Could a mass really hand like this (on a single washing line) in equilibrium with angle a different to angle b?


Well the follow-up question was to explain why the angles would be similar in value...
The answer to that would be that the jacket is likely to slide to the centre to maintain equilibria
Reply 10
Original post by notnek
Could a mass really hand like this (on a single washing line) in equilibrium with angle a different to angle b?


Ah, I see what you mean!
Reply 11
Original post by Someboady
Well the follow-up question was to explain why the angles would be similar in value...
The answer to that would be that the jacket is likely to slide to the centre to maintain equilibria

Ah that makes a bit more sense.

Where did you get this question from? I doubt it would be in a real M1 paper. The angles and tension on either side will be the same for a question like this (probably).
Reply 12
Original post by notnek
Ah that makes a bit more sense.

Where did you get this question from? I doubt it would be in a real M1 paper. The angles and tension on either side will be the same for a question like this (probably).


Aye, I was doing the Solomon Paper C. Agreed, the questions are quite unorthodox tbh
Reply 13
Original post by Zacken
Ah, I see what you mean!


Just the one more question xD

What generally happens when a string breaks in Mechanics?
E.g.
On a rough slope where a particle attached to the string is accelerating, apparently it is "catapulted".
What if it wasn't accelerating and moving at constant velocity i.e. in equillibrium, would it still be "catapulted"?
The tension in the string would disappear but what would happen to the friction at the point where the string breaks (instantly), do we assume friction acts in the opposite direction or do we just assume friction is zero(i think its this)?

How about when you have a pulley with two particles of different masses attached and one accelerates. Suddenly the string breaks. Tension would disappear but again, is the particle somehow catapulted upwards and is acting under g?

I believe my assumptions are correct but I'm not sure. Sorry for continuously bothering you :P :colondollar:
Reply 14
Original post by Someboady
Just the one more question xD

What generally happens when a string breaks in Mechanics?
E.g.
On a rough slope where a particle attached to the string is accelerating, apparently it is "catapulted".
What if it wasn't accelerating and moving at constant velocity i.e. in equillibrium, would it still be "catapulted"?
The tension in the string would disappear but what would happen to the friction at the point where the string breaks (instantly), do we assume friction acts in the opposite direction or do we just assume friction is zero(i think its this)?

How about when you have a pulley with two particles of different masses attached and one accelerates. Suddenly the string breaks. Tension would disappear but again, is the particle somehow catapulted upwards and is acting under g?

I believe my assumptions are correct but I'm not sure. Sorry for continuously bothering you :P :colondollar:


Afaik, there is no catapulting or anything, when the string breaks it just means the tension becomes 0 so the only force acting on the particle is the friction which causes deceleration - this is for both pulleys and inclined planes.

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