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    (Original post by 2776)
    And so since F=uR

    We know F and we know R.

    and so u=F/R

    and I get u = 0.349
    Which does not correspond with the u that we know already
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    (Original post by 2776)
    And so since F=uR

    We know F and we know R.

    and so u=F/R

    and I get u = 0.349
    hmmmm.....i still think μ should be constant....0.364.
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    (Original post by king of swords)
    hmmmm.....i still think μ should be constant....0.364.
    If yours comes out thinking that μ is less i guess that is a good thing because it really means that F has not reached its highest value yet.
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    (Original post by king of swords)
    hmmmm.....i still think μ should be constant....0.364.
    Yes, we need to prove that u is constant.
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    (Original post by 2776)
    Yes, we need to prove that u is constant.
    No you don't.....μ just IS constant....it's that being constant that you should use to prove the system is in equilibrium.
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    (Original post by king of swords)
    No you don't.....μ just IS constant....it's that being constant that you should use to prove the system is in equilibrium.
    so what else should I do to my calculation?
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    (Original post by 2776)
    so what else should I do to my calculation?
    well i'll just show you my confusion of it all:
    if in equilibrium:
    F + 250gsin20 = 2000cos20
    R + 2000sin20 = 250gcos20
    therefore FMAX = μR = μ(250gcos20 - 2000sin20) = tan20(250gcos20 - 2000sin20) = 589.0N

    Also F = 2000cos20 - 250gsin20 = 1041.4N!!!! which exceeds FMAX...therefore i cannot see how this can be in equilibrium
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    (Original post by king of swords)
    well i'll just show you my confusion of it all:
    if in equilibrium:
    F + 250gsin20 = 2000cos20
    R + 2000sin20 = 250gcos20
    therefore FMAX = μR = μ(250gcos20 - 2000sin20) = tan20(250gcos20 - 2000sin20) = 589.0N

    Also F = 2000cos20 - 250gsin20 = 1041.4N!!!! which exceeds FMAX...therefore i cannot see how this can be in equilibrium
    I think your 2 equations are wrong. See me 2 equations (latest one) on page 2. That is the ocrrect one
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    R- (250gcos20+2000sin20)=0

    F+ (2000cos20-250gsin20)=0
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    (Original post by 2776)
    R- (250gcos20+2000sin20)=0

    F+ (2000cos20-250gsin20)=0
    I'm guessing you have F going up the plane? and yes i've messed up the equation in R.
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    (Original post by king of swords)
    I'm guessing you have F going up the plane? and yes i've messed up the equation in R.
    That is correct. But then it is totally wrong. Cause I just checked, I get u= -0.348.

    Which is totally wrong.
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    Cause F= 2986.287...

    R= -1041.4...
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    (Original post by 2776)
    That is correct. But then it is totally wrong. Cause I just checked, I get u= -0.348.

    Which is totally wrong.
    Well F should be going down the plane if the horizontal 2000N force is going "into" the plane. Remember, F changes to work against motion.
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    (Original post by king of swords)
    Well F should be going down the plane if the horizontal 2000N force is going "into" the plane. Remember, F changes to work against motion.
    done it....wait a minute
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    (Original post by king of swords)
    Well F should be going down the plane if the horizontal 2000N force is going "into" the plane. Remember, F changes to work against motion.
    Thats clever, yes I get it now. As 2000N is used to push up the plane, F should oppose action.

    I mean this question is a killer isnt it? Its in OCR 2001 test. Did u do that?
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    (Original post by king of swords)
    done it....wait a minute
    F does go down the plane:
    if in equilibrium:
    F = 2000cos20 - 250gsin20 = 1041.4N
    R = 2000sin20 + 250gcos20
    so FMAX = μR = tan20(2000sin20 + 250gcos20) = 1086.9N which means that FMAX is greater than the actual value of F in this circumstance which means the system is in equilibrium.....sorry about that.
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    (Original post by king of swords)
    F does go down the plane:
    if in equilibrium:
    F = 2000cos20 - 250gsin20 = 1041.4N
    R = 2000sin20 + 250gcos20
    so FMAX = μR = tan20(2000sin20 + 250gcos20) = 1086.9N which means that FMAX is greater than the actual value of F in this circumstance which means the system is in equilibrium.....sorry about that.
    Right another question I'm stuck on:

    particle A: 0.15kg (2m/s towards B)
    particle B: 0.2 kg (stationary)

    The coeffient of friction between A and surface is 0.05. a is brought to rest after collision. What is speed of B?

    If it was a smooth surface, I have calculated that B moves at 3m/s away.
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    (Original post by 2776)
    Right another question I'm stuck on:

    particle A: 0.15kg (2m/s towards B)
    particle B: 0.2 kg (stationary)

    The coeffient of friction between A and surface is 0.05. a is brought to rest after collision. What is speed of B?

    If it was a smooth surface, I have calculated that B moves at 3m/s away.
    The particles are 4 m away form each other. I reckon that I should find the ending velocity i.e. u (2).
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    (Original post by 2776)
    Right another question I'm stuck on:

    particle A: 0.15kg (2m/s towards B)
    particle B: 0.2 kg (stationary)

    The coeffient of friction between A and surface is 0.05. a is brought to rest after collision. What is speed of B?

    If it was a smooth surface, I have calculated that B moves at 3m/s away.
    It would help if you knew whether the collision was inelastic or elastic.
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    (Original post by king of swords)
    It would help if you knew whether the collision was inelastic or elastic.
    Im not too sure, I think it wants me to use the momentum thing.

    Anyway, the speed of B is 1.5m/s after collision on a smooth plane(not 3m/s)

    Doesnt give the info, whether it is selastic or not.
 
 
 
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