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    I know that this should be a simple question, but I really am quite stuck!

    (For those who have the Heinemann Textbooks this is Ex 5F Q14)

    Particle A of mass 1kg is at rest 0.2m from the edge of a smooth horizontal 0.8m high table. It is connected by a light inextensible string of length 0.7m to a particle B of mass 0.5kg. Particle B is initially at rest at the edge of the table closest to A but then falls off. Assuming B's initial horizontal velocity to be zero, find the speed with which A begins to move.

    (The answer at the back of the book is 1.04m/s)

    Thanks
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    v² = u² + 2as
    vb² = 2x9.8x(0.7-0.2)
    vb = 3.13

    maua + mbub = mava + mbvb

    so as the 2 particles will move at the same velocity after the string becomes tight .. then:

    maua + mbub = mav + mbv

    0 + 0.5x3.13 = 1xv + 0.5xv
    1.57 = 1.5 v
    v = 1.04 ms-1
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    I have one more question:

    (M1 Ex4E Q6b)

    A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a force Y acting at an angle of 45 degrees to the horizontal. The coefficient of friction between the body and the plane is 0.2. Find Y when the body is on the point of slipping down the plane.

    I have tried the following method to solve this problem:

    Resolving up the plane gives:
    F - mgsin20 - Ycos65 = 0
    F - 19.6sin20 - Ycos65 = 0

    Resolving perpendicular to the plane gives:
    R - mgcos20 + Ysin65 = 0
    R - 19.6cos20 + Ysin65 = 0

    As F=0.2R,

    0.2(19.6cos20 - Ysin65) - 19.6sin20 - Ycos65 = 0
    Y(cos65 + 0.2sin 65) = 3.92cos20 - 19.6sin 20

    Therefore Y=-6.15N (3sf.)

    However, the back of the book says that is 3.67N

    Could anyone see where I went wrong, and what the correct method is?

    Thanks again
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    you've used 65 degrees instead of the stated 45.
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    (Original post by Goldenratio)
    you've used 65 degrees instead of the stated 45.
    I did that purposely. Y = 20 + 45 = 65 degrees away from the perpendicular distance of R (I think)

    If you use 45 degrees instead, the answer comes out to be -5.00N, which is still incorrect.

    Perhaps a diagram would help
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    Anyone else?
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    (Original post by ebinsc)
    I have one more question:

    (M1 Ex4E Q6b)

    A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a force Y acting at an angle of 45 degrees to the horizontal. The coefficient of friction between the body and the plane is 0.2. Find Y when the body is on the point of slipping down the plane.

    I have tried the following method to solve this problem:

    Resolving up the plane gives:
    F - mgsin20 - Ycos65 = 0
    F - 19.6sin20 - Ycos65 = 0

    Resolving perpendicular to the plane gives:
    R - mgcos20 + Ysin65 = 0
    R - 19.6cos20 + Ysin65 = 0

    As F=0.2R,

    0.2(19.6cos20 - Ysin65) - 19.6sin20 - Ycos65 = 0
    Y(cos65 + 0.2sin 65) = 3.92cos20 - 19.6sin 20

    Therefore Y=-6.15N (3sf.)

    However, the back of the book says that is 3.67N

    Could anyone see where I went wrong, and what the correct method is?

    Thanks again
    Limiting equilibrium
    => Friction = coefficient of friction x normal reaction
    = 0.2R

    PERPENDICULAR TO PLANE: R = 2gcos20+Ycos65
    PARALLEL TO PLANE: 0.2R = 2gsin20+Ycos25
    0.2(2gcos20+Ycos65) = 2gsin20+Ycos25
    Y(cos25-0.2cos65) = 2gsin20-0.4gcos20
    Y = (2gsin20-0.4gcos20)/(cos25-0.2cos65)
    Y = (19.6sin20-3.92cos20)/(cos25-0.2cos65)
    Y = 3.67N
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    (Original post by Widowmaker)
    Limiting equilibrium
    => Friction = coefficient of friction x normal reaction
    = 0.2R

    PERPENDICULAR TO PLANE: R = 2gcos20+Ycos65
    PARALLEL TO PLANE: 0.2R = 2gsin20+Ycos25
    0.2(2gcos20+Ycos65) = 2gsin20+Ycos25
    Y(cos25-0.2cos65) = 2gsin20-0.4gcos20
    Y = (2gsin20-0.4gcos20)/(cos25-0.2cos65)
    Y = (19.6sin20-3.92cos20)/(cos25-0.2cos65)
    Y = 3.67N
    Thanks
 
 
 
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