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M1: pulley question on inclined planes Watch

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    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!
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    (Original post by tessa.lin)
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    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.
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    (Original post by tessa.lin)
    Name:  1453751667380-571705251.jpg
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    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).
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    (Original post by tessa.lin)
    Name:  1453751667380-571705251.jpg
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    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?
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    (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?
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    (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.
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    (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?
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    (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.
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    (Original post by notnek)
    Sorry I meant particle P, not Q. P is on a rough surface and Q is on a smooth surface.
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    Is this correct?
    Can the m from both sides of the equations be crossed out?
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    (Original post by tessa.lin)
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    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.
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    (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
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    (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.
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    (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
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    (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.
 
 
 
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