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A smooth ring R of weight W N is threaded on a light inextensible string. The ends of the string are attached to fixed points A and B, where A is vertically above B. A horizontal force of magnitude P N acts on R. The system is in equilibrium with the string taut; AR makes an angle a with the downward vertical and BR makes an angle b with the upward vertical (see diagram below).
A
| -
|a -
| -
| - R --------->P N
| b -
|-
B
It is given that when P = 14, AR = 0.4 m, BR = 0.3 m and the perpendicular distance of R from the vertical line AB is 0.24 m.
Find:
(a) the tension in the string;
(b) the value of W.
For the case when P = 0,
(c) describe the position of R,
(d) state the tension in the string.
It's the last 2 parts that I'm struggling with.
Thanks for any help.
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hi jimi70
If there is no horizontal force, there's nothing to keep the string stretched out to the side.
So the weight will just hang vertically down below A.
The total length of the string is 0.7 and AB is 0.24.
So there must be 0.46 m to account for.
So the weight will hang below B at a distance PB = 0.23 m
The weight is supported by two pieces of string, each with tension, T.
So 2T = W.
Hope that helps.
Bob
ps. What answers did you get for (a) and (b) ?
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks bob.
for (a) T sin a + T sin b = 14
sin a = 0.24/0.4 = 0.6
sin b = 0.24/0.3 = 0.8
so: 1.4T = 14 and T = 10 N
T cos a - T cos b = W
cos a = 0.8
cos b = 0.6
so: 8 - 6 = W and W = 2 N
'The total length of the string is 0.7 and AB is 0.24.
So there must be 0.46 m to account for.'
Sorry, it's probably my awful diagram, but AB must be 0.5 m. That means that there will be 0.2 m to account for. Does that mean that R will hang 0.1 m below B?
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hi jimi70
Just re-read the problem. I was taking AB = 0.24 m. That makes the problem quite a bit harder as there is no right angle you can use.
Now I've looked again, I'm happy with what you have done.
sin a = 0.6 and sin b = 0.8 => the angle at R is 90 and so AB = 0.5 m
So in part (c) R is 0.1 below B as you say.
Bob
ps. you can download 'geogebra' for free and make pretty good diagrams.
bobbym gave a number of sites for this at
http://www.mathisfunforum.com/viewtopic.php?id=17452
Then make a screen shot and save as a .gif
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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bob
Sonething just occurred to me. If the string is not now taut, does that mean that we cannot tell exactly how far the string will hang below B?
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The ring has weight so it will just hang down under gravity, with the string slipping through the ring until both strands are equally taut.
see picture
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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