M1: pulley question on inclined planes

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!

Reply 1

Zacken

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.

Reply 2

Notnek

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).

Reply 3

TeeEm

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?

Reply 4

username1391690

Original post by Zacken

Original post by notnek

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?

Reply 5

Notnek

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.

Reply 6

username1391690

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?

Reply 7

Notnek

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.

Reply 8

username1391690

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?

Reply 9

Notnek

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.

Reply 10

username1391690

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

Reply 11

Notnek

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.

Reply 12

username1391690

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

Reply 13

Notnek

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