# Need some help on this M1 pulley question

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#1

I have solved part A-D of this question, the answers are

A: Acceleration of P while it is in motion = 4.9 ms-2

B: Tension in the string = 1.5 N

C: Mass of m = 0.10 kg

D: Greatest height of Q above the ground = 0.50 m

But really stuck on part E and F... can anyone lend me a helping hand please?
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3 years ago
#2
For part E, the pulley is not moving, so must have zero net force acting on it. What is acting downwards on the pulley?

For part F, particle P is supported by the horizontal surface and particle Q is moving freely under the influence of gravity (i.e. isn't supported by anything).
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#3
(Original post by old_engineer)
For part E, the pulley is not moving, so must have zero net force acting on it. What is acting downwards on the pulley?

For part F, particle P is supported by the horizontal surface and particle Q is moving freely under the influence of gravity (i.e. isn't supported by anything).
For E, I think there are weight of the pulley and weight of the two particles acting downwards. But when I tried to type in T = (0.5+0.3+0.1) x 9.8 = 8.8 N it says incorrect...?
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3 years ago
#4
(Original post by ML8020)
For E, I think there are weight of the pulley and weight of the two particles acting downwards. But when I tried to type in T = (0.5+0.3+0.1) x 9.8 = 8.8 N it says incorrect...?
The downward force of the string on the pulley consists of two lots of string tension. (Using the masses of P and Q would only be correct if they were being supported in equilibrium).
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#5
(Original post by old_engineer)
The downward force of the string on the pulley consists of two lots of string tension. (Using the masses of P and Q would only be correct if they were being supported in equilibrium).
I can't quite understand what I should do instead... would you mind giving me a bigger hint? Sorry for the confusion.
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3 years ago
#6
You've worked out the string tension in part B. Each side of the string is pulling down on the pulley with that amount of force.
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#7
(Original post by old_engineer)
You've worked out the string tension in part B. Each side of the string is pulling down on the pulley with that amount of force.
Oh yes got part E.
It's T = 0.5 x 9.81 + 1.5 +1.5 = 7.9 N
Thanks for the guidance

Only part F to go.
Now P has landed so the string is no longer taut, meaning there is no tension on both particles?
As a result T = 0.5 x 9.81 + 0.1 x 9.81 (weight of Q) = 5.9 N?
But it's incorrect...
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3 years ago
#8
(Original post by ML8020)
Oh yes got part E.
It's T = 0.5 x 9.81 + 1.5 +1.5 = 7.9 N
Thanks for the guidance

Only part F to go.
Now P has landed so the string is no longer taut, meaning there is no tension on both particles?
As a result T = 0.5 x 9.81 + 0.1 x 9.81 (weight of Q) = 5.9 N?
But it's incorrect...
Q is moving upwards freely under the influence of gravity. It is not being supported by the string (the string is slack). Therefore the pulley is not supporting Q, and the weight of Q is not contributing to the tension in the chain.
1
#9
(Original post by old_engineer)
Q is moving upwards freely under the influence of gravity. It is not being supported by the string (the string is slack). Therefore the pulley is not supporting Q, and the weight of Q is not contributing to the tension in the chain.
Oh yeah finally got it. How come I didn't realise when the string is slack BOTH particles no longer contribute to the tension in the chain. The answer is simply 0.5 x 9.81 = 4.9 N

Thanks a lot for your time! I should have noticed this way earlier
(Just the Board one to go now)
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