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AS maths f=ma help

I'm so bad at these type questions could someone just explain how to do them?

Two particles one of madd 4kg and the other of mass 3kg are joined by a light inextensible string which passes over a smooth fixed pulley. Find the acceleration of the particles and the tension in the string in terms of g.

Thank you!

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Original post by Middlechild123
I'm so bad at these type questions could someone just explain how to do them?

Two particles one of madd 4kg and the other of mass 3kg are joined by a light inextensible string which passes over a smooth fixed pulley. Find the acceleration of the particles and the tension in the string in terms of g.

Thank you!


Draw a diagram, mark forces in particles. Which one moves down?

Post answer to that and I'll help with next step.
Original post by Muttley79
Draw a diagram, mark forces in particles. Which one moves down?

Post answer to that and I'll help with next step.

Ok sorry this took a while, my photo won't upload but I drew basically a circle with two strings coming off either side of it donwards which have weights on the end and each one had an arrow pointing up saying T (along the rope) and each one had an arrow from the weight down saying W.
Reply 3
OK, so there is a Force = mg on each side of the pulley, 4g N for one mass and 3g N for the other, That gives you a resultant force of.......?

That force is accelerating BOTH masses, so use your resultant force with the total mass to find the acceleration. Common sense should tell you in which direction....

Now to find the tension, look at one of the masses, let's say the 4kg mass. Its weight is 4g. The resultant force on it you can calculate using F=ma but this time the m is just the 4kg and the a is the one you've just calculated.

Now you know the resultant force, you know the weight, the only other force acting on it is the Tension so you shouldbe able to deduce that.

Hope this helps.
Original post by phys981
OK, so there is a Force = mg on each side of the pulley, 4g N for one mass and 3g N for the other, That gives you a resultant force of.......?

That force is accelerating BOTH masses, so use your resultant force with the total mass to find the acceleration. Common sense should tell you in which direction....

Now to find the tension, look at one of the masses, let's say the 4kg mass. Its weight is 4g. The resultant force on it you can calculate using F=ma but this time the m is just the 4kg and the a is the one you've just calculated.

Now you know the resultant force, you know the weight, the only other force acting on it is the Tension so you shouldbe able to deduce that.

Hope this helps.

Ok thank you that's really helpful, so the forces are 29.4 and 39.2 and do I add them together?
Reply 5
Look at which way they're acting, one is pulling down in one direction, the other is pulling down on the other side so in the other direction on the string. In other words, they're acting against each other, so the resultant is the difference between them, not the sum.


(This is logical if you think about it - clearly the one with more mass is going to accelerate downwards but its acceleration will be LESS than if the 3kg one wasn't there)
(edited 5 years ago)
Original post by Middlechild123
Ok sorry this took a while, my photo won't upload but I drew basically a circle with two strings coming off either side of it donwards which have weights on the end and each one had an arrow pointing up saying T (along the rope) and each one had an arrow from the weight down saying W.


OK which one moves down? Take that direction as positive.

Use F=ma on this weight, post your answer.
Original post by Muttley79
OK which one moves down? Take that direction as positive.

Use F=ma on this weight, post your answer.


But don't they both move in the same direction?
Original post by phys981
Look at which way they're acting, one is pulling down in one direction, the other is pulling down on the other side so in the other direction on the string. In other words, they're acting against each other, so the resultant is the difference between them, not the sum.


(This is logical if you think about it - clearly the one with more mass is going to accelerate downwards but its acceleration will be LESS than if the 3kg one wasn't there)

Oh right of course that makes sense thank you
Original post by phys981
OK, so there is a Force = mg on each side of the pulley, 4g N for one mass and 3g N for the other, That gives you a resultant force of.......?

That force is accelerating BOTH masses, so use your resultant force with the total mass to find the acceleration. Common sense should tell you in which direction....

Now to find the tension, look at one of the masses, let's say the 4kg mass. Its weight is 4g. The resultant force on it you can calculate using F=ma but this time the m is just the 4kg and the a is the one you've just calculated.

Now you know the resultant force, you know the weight, the only other force acting on it is the Tension so you shouldbe able to deduce that.

Hope this helps.


You are missing out key steps - you MUST set out two equations using F= ma first.
Original post by Middlechild123
Oh right of course that makes sense thank you


Please read my posts not the advice from the other poster.
Original post by Middlechild123
But don't they both move in the same direction?


Since they're connected to the same string, as one moves down, the other must move up.
Original post by Muttley79
You are missing out key steps - you MUST set out two equations using F= ma first.


So a=f/m and m is 7 because 3+4?
And 39.2-29.4=9.8 so 9.8 is the resultant force?
Original post by Middlechild123
So a=f/m and m is 7 because 3+4?
And 39.2-29.4=9.8 so 9.8 is the resultant force?


Set out your two equations:

First one on the 4kg mass

Using F=ma on 4kg gives:

4g - T = 4a

Now do the other one;
Original post by Muttley79
Set out your two equations:

First one on the 4kg mass

Using F=ma on 4kg gives:

4g - T = 4a

Now do the other one;


T-3g=3a?
Original post by Middlechild123
T-3g=3a?


Yes, now if you want the acceleration you need to add the two equations to eliminate T.

You must not short-cut this process ... you need to set up two simultaneous equations.
Original post by Muttley79
Yes, now if you want the acceleration you need to add the two equations to eliminate T.

You must not short-cut this process ... you need to set up two simultaneous equations.


So g=7a and therefore a=g/7? Then what? Thank you
Original post by Middlechild123
So g=7a and therefore a=g/7? Then what? Thank you


You have one solution of your simultaneous equations so substitute the value of a in to get T.

This method will work for all pulleys where both strings are vertical. If one particle is on a table the same approach works but friction might be involved if the table is rough.
Original post by Muttley79
You have one solution of your simultaneous equations so substitute the value of a in to get T.

This method will work for all pulleys where both strings are vertical. If one particle is on a table the same approach works but friction might be involved if the table is rough.


So,

a=g/7
a=9.8/7
a=1.4

4g-4a=T
(4x9.8)-(4x1.4)=T
T=33.6 with the string with the 4kg weight?
And it is accelerating down as it is heavier?

How do I do the 3kg one?
Original post by Middlechild123
So,

a=g/7
a=9.8/7
a=1.4

4g-4a=T
(4x9.8)-(4x1.4)=T
T=33.6 with the string with the 4kg weight?
And it is accelerating down as it is heavier?

How do I do the 3kg one?


Don't forget units on the tension.

You've done the 3kg one .... same acceleration throughout the system.

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