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Old May 8th 2009, 05:45 AM   #1
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String systems

Greetings. This has always been my least favorite topic, here are 2 questions in which I need assistance

1) A particle A, of mass m kg, rests on a smooth horizontal table. It is connected by a smooth, light, inextensible string which passes over a smooth, light, fixed pulley to a 2nd particle B, of mass 2kg, which hangs freely under gravity. The system starts from rest with A at a distance of 1 meter from the pulley.
Calculate the acceleration of A.

2) A particle of weight W is attached to two light inextensible strings each of length 26 cm. The other ends of the string are attached to two points on the same horizontal level 48 cm apart. Find the tension in each of the strings in terms of W.



Thanks

Last edited by Beef_PiE; May 8th 2009 at 05:50 AM.
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Old May 8th 2009, 08:22 AM   #2
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hi

hi

1)

For object at A: $\displaystyle T = m_{1}a $
For object at B: $\displaystyle mg-T=m_{2}a $

Eliminating T and solving for a gives us: $\displaystyle a = \frac{mg}{(m_{1}+m_{2})} $

In 2) I am not sure exactly what you mean. Do you mean that the two strings and horizontal make up a triangle, with the object hanging freely?
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Old May 8th 2009, 06:04 PM   #3
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We meet again...Twig. For 1), I believe that I can safely assume that acceleration is expressed in terms of the mass of A as it is currently unknown? I got the same answer but was unsure as it had an unknown in it.

For 2) I'm not exactly sure what I mean either... I've copied the question word for word. There was no accompanying diagram. But I assume it's safe to assume that the two strings and the horizontal form a triangle, with the object hanging freely.
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Old May 9th 2009, 01:25 AM   #4
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ok, then I guess the strings make up an isoceles triangle. So the tension in them will be the same.

You need to put down some equilibirum equations.

$\displaystyle \sum F_{x} = 0 \mbox{ and }\sum F_{y}=0 $ .

So that the net force in both x and y directions are zero. Otherwise the object would be subject to an acceleration $\displaystyle a = \frac{F}{m} $ .
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Old May 9th 2009, 02:10 AM   #5
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2) A particle of weight W is attached to two light inextensible strings each of length 26 cm. The other ends of the string are attached to two points on the same horizontal level 48 cm apart. Find the tension in each of the strings in terms of W.

Because both the strings are 26 cm, and the symmetry of the problem, the weight will hang at a point 10 cm below the midpoint of the 48 cm.
26^2 - 24^2 = 100.
If @ = angle each string makes with the vertical, we have cos@ = 10/26
Let T be the tension in each string.

For equilibrium, W = 2Tcos@ = (20/26) T or T = 1.3 W

The sine components of T cancel horizontally as they are in opposite directions.

Note that the tension is more than the weight itself !
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