Tut
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#1
Report Thread starter 13 years ago
#1
Hey guys,

I've been trying to revise the connected particles chapter and i am getting nowhere!.. I'm just so confused about everything in this part of M1. Anybody got any tips or any help they could give. I have a question that i have been told if you can do, you do not need 2 worry about any connected particles questions because it has everything in there you need 2 solve. If anybody could answer it step-by-step for i would appreciate greatly.

A particle A of mass 0.8kg rests on a horizontal table and is attached to one end of a light inextensible string. The string passes over a small pulley P fixed at the edge of the table. The other end of the string is attached to a particle B of mass 1.2kg which hangs freely below the pulley. The system is released from rest with the string taut and with B at a height of 0.6m above the ground. In the subsequent motion A does not reach P before B reaches the ground. In an inital model of the situation, the table is assumed to be smooth.

(a) Find the tension in the string before B reaches the ground

(b) The time taken for B to reach the ground

In a refinement of the model, it is assumed that the table is rough and that the coefficient of friction between A and the table is 1/5.

(c) Find the time taken for B to hit the ground.

I know this is a lot to ask but any help is very much appreciated.
Thanx

Tut
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#2
Report 13 years ago
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1. Consider f = ma for both particles. Remember that the only force accelerating this system, initially, is the 1.2g which B will have, due to its weight.

So, for A, you get T = 0.8a
and for B, you get 1.2g - T = 1.2a

(Note the order of the T and the g. The tension is the resistive force for B and the main pulling force for the object A, and assuming that the system moves downwards you have to have them in this order.)

Now simply add the two equations, T cancels. 1.2g = 2a, a => 3/5 g.

Subbing back into either of the equations, T = 0.8(3/5 g)
Solve for T. T => 12/25 g. (or 0.48g)

2. Using s = ut + (1/2)at^2. Simply let u = 0, a (as we already know) = 3/5 g. and s = 0.6. And solve for t. t => 0.45.

3. We just need to reconsider the situation for B now. Taking into account the fact that its now T - µ(0.8g) = 0.8a.
F = µR, F = (1/5) * (0.8g) = 4/25 g

T - (4/25)g = 0.8a (new equation for A)
1.2g - T = 1.2a

Again, just add to eliminate T.

1.04g = 2a
a => 0.52g.

Finally, using s = ut + (1/2)at^2 again, using u = 0, s = 0.6m and a = 0.52 g.

Solve to find that t => 0.49s.
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Tut
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#3
Report Thread starter 13 years ago
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Thanx.. I understand a lot better than i did before!! Appreciate your help :-)

Tut
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Simply_The_best
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a box A of mass 0.8kg rest on a rough horizontal table and is attached to one end of a light inextensible string. the string passes over a smooth pulley fixed at the edge of the table. The other end of the string is attached to a sphere B of the edge of the table. The other end of the string is attached to a sphere B of mass 1.2kg, which hangs freely below the pulley. The magnitude of the frictional force between A and the table is FN. The system is released from rest with the string taut. After release, B DESCENDS A DISTANCE OF 0.9M IN 0.8S . Modelling A and B as particles, calculate: (a) the acceleration of B (b) the tension in the string (c) the value of F (d) find the total distance travelled by A
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ghostwalker
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#5
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(Original post by Simply_The_best)
..
This thread is over 11 years old!

As you're new to TSR, you may not be aware of the rules for the maths forum. Please see the "Posting Guidelines" thread at the top of the forum, and when you're ready start a new thread.
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