Pulleys F=ma

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A0W0N
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#1
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#1
I have this question which i got wrong can someone explain my mistake?
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A0W0N
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#2
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i put in the values and it was incorrect??
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RDKGames
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#3
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(Original post by A0W0N)
i put in the values and it was incorrect??
Not correct.

You should consider both masses separately and construct two equations in two variables; T for tension in the string, and a for acceleration of both particles.

Hence solve these simultaneously.
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A0W0N
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#4
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(Original post by RDKGames)
Not correct.

You should consider both masses separately and construct two equations in two variables; T for tension in the string, and a for acceleration of both particles.

Hence solve these simultaneously.
So do t=5*9.8
And t=7*9.8
And set them equal to each other?
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davros
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#5
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(Original post by A0W0N)
So do t=5*9.8
And t=7*9.8
And set them equal to each other?
Have you seen examples of pulley systems in your textbook before?

You need to consider the forces acting on each object in terms of its weight and the tension in the string, then set up equations of motion for each one which you can solve in the usual way for simultaneous equations.
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A0W0N
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#6
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(Original post by davros)
Have you seen examples of pulley systems in your textbook before?

You need to consider the forces acting on each object in terms of its weight and the tension in the string, then set up equations of motion for each one which you can solve in the usual way for simultaneous equations.
I have but it was a while back do they both add up to equal t? I can't remember
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simon0
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#7
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Your diagram is incomplete as you need to add the weight for each mass!

See which mass will go down and which mass will go up.

Then resolve for each mass separately.

I will give first one:

Resolving for 7 kg mass downwards:  \displaystyle 7g - T = 7a,
where T is the string tension and a is the acceleration of the 7 kg mass.

Can you:
- Tell me which mass will go down?
- Resolve for the 5 kg mass?
Last edited by simon0; 2 years ago
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A0W0N
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#8
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(Original post by simon0)
Your diagram is incomplete as you need to add the weight for each mass!

See which mass will go down and which mass will go up.

Then resolve for each mass separately.

I will give first one:

Resolving for 7 kg mass downwards:  \displaystyle 7g - T = 7a,
where T is the string tension and a is the acceleration of the 7 kg mass.

Can you:
- Tell me which mass will go down?
- Resolve for the 5 kg mass?
7kg mass will go down, so t is negative so for the 5kg it will go up meaning it will be: T-5g=5a?
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username4197954
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#9
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I don't know if this helps, but this is how I like to draw the diagrams for pulleys. I've labelled the masses B and C - this helps me add the forces. I consider the direction the masses will move and then add the acceleration label in that direction. There will be tension in the string, as added. Then it's just a case of resolving the forces for each mass using F=ma. Then simutaneously solving the 2 equations to get the acceleration (a).

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A0W0N
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#10
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I got it now thanks, I hate to ask but im stuck on another question
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username4197954
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#11
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#11
Clue .... don't forget that mass exerts a force of weight.

Weight = mass x gravity (W=mg)

E.g. a mass of 3kg exerts a force of 3g N (or 3 x 9.8 = 29.4 N, where g=9.8 N/kg)
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simon0
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#12
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#12
(Original post by A0W0N)
I got it now thanks, I hate to ask but im stuck on another question
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Again, where is the weight?

Add the force of weight (weight of the bottle) and resolve the forces so to substitute into the F=ma, formula.
Last edited by simon0; 2 years ago
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A0W0N
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I don't understand why this is wrong, and i'd like to ask are t only equal to each other in equilibrium?
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username4197954
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#14
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#14
You've got the force of the 3kg mass acting in the wrong direction. Weight = mass x gravity (W=mg), so think about the direction of the force of gravity.

Also, there is one other force that is missing. If I said "Reaction" would that help? (trying not to give you the answer as I've been told off for that before, lol!)
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A0W0N
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#15
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(Original post by Bexr)
You've got the force of the 3kg mass acting in the wrong direction. Weight = mass x gravity (W=mg), so think about the direction of the force of gravity.

Also, there is one other force that is missing. If I said "Reaction" would that help? (trying not to give you the answer as I've been told off for that before, lol!)
umm is the reaction force upwards from the 3kg mass?
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simon0
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#16
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(Original post by A0W0N)
umm is the reaction force upwards from the 3kg mass?
If the diagram is correct, yes.
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A0W0N
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#17
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(Original post by simon0)
If the diagram is correct, yes.
How will that affect the horizontal force, I don’t understand
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davros
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#18
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#18
(Original post by A0W0N)
How will that affect the horizontal force, I don’t understand
It won't in this case because the table is smooth. The reaction force balances the weight of the 3 kg mass acting downwards so it is neither "lifting off" nor burrowing through the table. If the table were rough then the reaction force would be involved in a friction calculation.

I think it was only mentioned for completeness i.e. to check that you were aware of all the forces acting on the mass.
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A0W0N
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#19
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#19
(Original post by davros)
It won't in this case because the table is smooth. The reaction force balances the weight of the 3 kg mass acting downwards so it is neither "lifting off" nor burrowing through the table. If the table were rough then the reaction force would be involved in a friction calculation.

I think it was only mentioned for completeness i.e. to check that you were aware of all the forces acting on the mass.
Ok thanks how do i find the acceleration though?
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davros
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#20
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#20
(Original post by A0W0N)
Ok thanks how do i find the acceleration though?
OK think about the 3 kg mass by itself. It's being pulled towards the pulley by a force T (tension in the string) and it accelerates towards the pulley with acceleration 'a', So how would you find 'a' using the normal Newton's 2nd law?
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