You are Here: Home >< Maths

# M1: pulley question on inclined planes watch

1. Is my diagram incorrect? Im not sur3 of mgcos45 because i turned the diagram around to treat P separately and thus the inclined angle was 45°.

Any input would be appreciated!
2. (Original post by tessa.lin)

Is my diagram incorrect? Im not sur3 of mgcos45 because i turned the diagram around to treat P separately and thus the inclined angle was 45°.

Any input would be appreciated!
Your diagram is fine, but mg cos 45 is incorrect. It should be another angle. Try drawing a line from a pulley to the base and separating it into two different inclined planes to help you get a feel for it.
3. (Original post by tessa.lin)

Is my diagram incorrect? Im not sur3 of mgcos45 because i turned the diagram around to treat P separately and thus the inclined angle was 45°.

Any input would be appreciated!
I'm not sure where you got 45 from.

if you just consider the right-hand-side of the diagram then this is just like a normal inclined slope question. The force component perpendicular to the slope is mgcos(60).
4. (Original post by tessa.lin)

Is my diagram incorrect? Im not sur3 of mgcos45 because i turned the diagram around to treat P separately and thus the inclined angle was 45°.

Any input would be appreciated!
shouldn't that be 60° on the right, for both sine and cosine?
5. (Original post by Zacken)
Your diagram is fine, but mg cos 45 is incorrect. It should be another angle. Try drawing a line from a pulley to the base and separating it into two different inclined planes to help you get a feel for it.

(Original post by notnek)
I'm not sure where you got 45 from.

if you just consider the right-hand-side of the diagram then this is just like a normal inclined slope question. The force component perpendicular to the slope is mgcos(60).
(Original post by TeeEm)
shouldn't that be 60° on the right, for both sine and cosine?
Thank you, Ive corrected myself now. To find tha acceleration do I use the reaction force at all? What do I start by resolving for?
6. (Original post by tessa.lin)
Thank you, Ive corrected myself now. To find tha acceleration do I use the reaction force at all? What do I start by resolving for?
You will need to find the reaction for particle Q so you can get friction.

Try your best and post any working (doesn't matter if it is wrong). We can correct you if you make a mistake.
7. (Original post by notnek)
You will need to find the reaction for particle Q so you can get friction.

Try your best and post any working (doesn't matter if it is wrong). We can correct you if you make a mistake.
Oh i thought if the question said a smooth surface then we can assume there is no friction?
8. (Original post by tessa.lin)
Oh i thought if the question said a smooth surface then we can assume there is no friction?
Sorry I meant particle P, not Q. P is on a rough surface and Q is on a smooth surface.
9. (Original post by notnek)
Sorry I meant particle P, not Q. P is on a rough surface and Q is on a smooth surface.

Is this correct?
Can the m from both sides of the equations be crossed out?
10. (Original post by tessa.lin)

Is this correct?
Can the m from both sides of the equations be crossed out?
You're nearly there for P but you seem to have forgotten tension.
11. (Original post by notnek)
You're nearly there for P but you seem to have forgotten tension.
Oh right. So it would be:
Mgsin60-0.5 x mgcos60 - T=ma

And I can get T from Q by:
R (diagonally upwards) Q: T- mgsin30= ma
12. (Original post by tessa.lin)
Oh right. So it would be:
Mgsin60-0.5 x mgcos60 - T=ma

And I can get T from Q by:
R (diagonally upwards) Q: T- mgsin30= ma
That's correct.

Now add the two equatiions to find the acceleration.
13. (Original post by notnek)
That's correct.

Now add the two equatiions to find the acceleration.
I might just be being very silly right now but doing this:
Mgsin60 -0.5mgcos60 - mgsin30 = ma
And crossinf out the mg is giving me -2.0839... which is far from the right answer
14. (Original post by tessa.lin)
I might just be being very silly right now but doing this:
Mgsin60 -0.5mgcos60 - mgsin30 = ma
And crossinf out the mg is giving me -2.0839... which is far from the right answer
When adding the right-hand-sides of the equations : ma + ma = 2ma.

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: January 25, 2016
Today on TSR

### Are you living on a tight budget at uni?

From budgets to cutbacks...

### University open days

1. University of Edinburgh
Sat, 22 Sep '18
2. University of Exeter
Sat, 22 Sep '18
3. Loughborough University
Sat, 22 Sep '18
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