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    (Original post by king of swords)
    You have three forces at work...weight, reaction force and friction.
    Equilibrium of forces // to the plane give you:
    0 = 250gsin20 - Friction
    If it then gives that the equilibrium is limiting, therefore:

    F=uR, where R is the resultant force. So:

    250g sin 20=u R

    But how do I find R?

    Thanks
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    (Original post by 2776)
    If it then gives that the equilibrium is limiting, therefore:

    F=uR, where R is the resultant force. So:

    250g sin 20=u R

    But how do I find R?

    Thanks
    Equilibrium perpendicular to the plane gives:
    0 = 250gcos20 - R
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    (Original post by king of swords)
    Equilibrium perpendicular to the plane gives:
    0 = 250gcos20 - R
    How do you get that?
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    (Original post by 2776)
    How do you get that?
    Like i said:
    equilibrium perpendicular to the plane....the only forces perpendicular to the plane are R and 250gcos20...and they are in opposite directions....it is in equilibrium so those two forces are also equal.
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    (Original post by king of swords)
    Like i said:
    equilibrium perpendicular to the plane....the only forces perpendicular to the plane are R and 250gcos20...and they are in opposite directions....it is in equilibrium so those two forces are also equal.
    I get this:

    u= (250gsin20)/(250gcos20)
    u=0.364 (3sf)

    Do you get this as well?
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    (Original post by 2776)
    I get this:

    u= (250gsin20)/(250gcos20)
    u=0.364 (3sf)

    Do you get this as well?
    yes
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    (Original post by king of swords)
    yes
    If a 2000N force is used to push the crate horizontally, show that the crate is in equilibrium.

    Parallel: 2000cos20=250gsin20+250gcos20
    Perpendicular: 250gsin20+2000gcos20

    is this correct?
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    (Original post by 2776)
    If a 2000N force is used to push the crate horizontally, show that the crate is in equilibrium.

    Parallel: 2000cos20=250gsin20+250gcos20
    Perpendicular: 250gsin20+2000gcos20

    is this correct?
    parallel: yes...
    perpendicular...you haven't got an = sign anywhere :confused:
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    (Original post by king of swords)
    parallel: yes...
    perpendicular...you haven't got an = sign anywhere :confused:
    perpendicular: R=250gcos20+2000sin20
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    (Original post by king of swords)
    parallel: yes...
    perpendicular...you haven't got an = sign anywhere :confused:
    One other thing....you are making certain forces equal to others as though you already know it is already known to be in equilibrium...the idea is to prove it.
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    (Original post by 2776)
    perpendicular: R=250gcos20+2000sin20
    250gcos20=250gcos20+2000sin20

    But then how do i prove that they are in equilibrium?
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    (Original post by 2776)
    250gcos20=250gcos20+2000sin20

    But then how do i prove that they are in equilibrium?
    Well that equation doesn't work if you look at it (because R no longer equals 250gcos20)
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    (Original post by king of swords)
    Well that equation doesn't work if you look at it (because R no longer equals 250gcos20)
    2000cos20=F+250gsin20

    R=250gcos20+2000sin20
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    (Original post by king of swords)
    Well that equation doesn't work if you look at it (because R no longer equals 250gcos20)
    I don't see how it can be in equilibrium....for perpendicular equilibrium....R must equal 1618.2N so the maximum value of F can only be 589.0N....but for parallel equilibrium F needs to be 1041.4N :confused:
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    (Original post by king of swords)
    I don't see how it can be in equilibrium....for perpendicular equilibrium....R must equal 1618.2N so the maximum value of F can only be 589.0N....but for parallel equilibrium F needs to be 1041.4N :confused:
    do you get that?
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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 reckon as a new force is added, u have to find the above 2 equations. and find f and r.
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    (Original post by 2776)
    I reckon as a new force is added, u have to find the above 2 equations. and find f and r.
    Yes so do I.....but one thing that stays constant is μ...because the crate and the surface are still the same materials.
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    (Original post by 2776)
    I reckon as a new force is added, u have to find the above 2 equations. and find f and r.
    R= 2986

    F=1041
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    (Original post by 2776)
    R= 2986

    F=1041
    And so since F=uR

    We know F and we know R.

    and so u=F/R

    and I get u = 0.349
 
 
 
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