You are Here: Home >< Maths

# m1 gravity Watch

1. Particles P and Q of masses 2m and m are attatched to ends of a light inextensible string which passes over a smooth fixed pulley. They both hang at a distance of 2m above horizontal ground, the system is released from rest.

Find the magnitude of acceleration of the system.

When resolving to do this, why do you use mg and 2mg for the mass of the particles? I thought you just used the mass? or is it weight (as w = mg ) in f = ma?
2. (Original post by lollage123)

When resolving to do this, why do you use mg and 2mg for the mass of the particles?
you don't
3. I think you'd just have:

(1) 2mg - T = 2ma
(2) T - mg = ma

But I haven't done mechanics in a while
4. Find net force then use combined mass
5. Ignore wat i said
6. (Original post by L'Evil Fish)
But I haven't done mechanics in a while
.... then it's been too long.......

The force that's making the 2m mass accelerate is gravity
The force that's opposing the movement of P is T
the effective force = mass x acceleration
2mg - T = 2ma

The force that's making Q accelerate comes from the tension in the string
The force that's opposing that is gravity
T - mg = ma
7. Pok im not sure anymore is is (2m-m)g/(2m+m)= g/3 assuming gravity = 10 it is 3.3333333333333333 ect
8. (Original post by gdunne42)
.... then it's been too long.......

The force that's making the 2m mass accelerate is gravity
The force that's opposing the movement of P is T
the effective force = mass x acceleration
2mg - T = 2ma

The force that's making Q accelerate comes from the tension in the string
The force that's opposing that is gravity
T - mg = ma
I left the g out

My apologies... I'll edit
9. (Original post by lollage123)
Particles P and Q of masses 2m and m are attatched to ends of a light inextensible string which passes over a smooth fixed pulley. They both hang at a distance of 2m above horizontal ground, the system is released from rest.

Find the magnitude of acceleration of the system.

When resolving to do this, why do you use mg and 2mg for the mass of the particles? I thought you just used the mass? or is it weight (as w = mg ) in f = ma?
The mass of each of the masses is 2m and m, right?
You do use F=ma, and in fact you have both, F and m, and you're looking for a.
the F is 2mg and mg, respectively.
Remember that the weight of a particle is:
• The force that particle has towards the Earth
• The force caused by acceleration due to gravity (
• In Newtons

Sub in your values of F.

1)
2)

Hopefully you can see a common factor that you can eliminate, then equate the equations to eachother and find a solution.

edit:
sorry I forgot about T and i'm far too tired to even be attempting to help anyone...
I hope this helps in any case
10. Ive seem to have done it in a much simpler way but got same answer as gdunne g/3 would i receive less marks
11. The tension bit is not need in my solution as it is inextensible making the system essentially a combined fixed object being moved
12. (Original post by chapmouse)
The mass of each of the masses is 2m and m, right?
You do use F=ma, and in fact you have both, F and m, and you're looking for a.
the F is 2mg and mg, respectively.
Remember that the weight of a particle is:
• The force that particle has towards the Earth
• The force caused by acceleration due to gravity (
• In Newtons

Sub in your values of F.

1)
2)

Spoiler:
Show
T = ma + mg

2mg - ma + mg = ma
3mg - 2ma = 0
m(3g-2a) = 0
So I get a to be 3g/2
^

Where have I gone wrong?

Posted from TSR Mobile
13. Okay thanks guys.
Knowing that the speed of P as it hits the ground is 3.6ms-1, and given that the particle q does not reach the pulley, how do you work out the height q reaches above the ground?
14. Use F=ma to calculate the acceleration of Q. Now you should be able to use suvat to calculate the height q reaches. Remember that the final velocity of q will be 0 because it will be momentarily stationary.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 7, 2013
Today on TSR

### Anxious about my Oxford offer

What should I do?

### Am I doomed because I messed up my mocks?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE