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    1) A plank AE, of length 6m and weight 100N, rests in a horizontal position on supports at B and D, where AB = 1m and DE = 1.5m. A child of weight 145N stands at C, the midpoint of AE. The plank is modelled as a uniform rod and the child and plank are in equilibrium.

    Calculate the magnitude of the force exerted by the support on the plank at B.

    Okay so R_B + R_D = 245
    Taking clockwise moments about B: M(B) = (245 x 2) - 4.5R_D = 0
    490 - 4.5R_D = 0 therefore R_D = 108.9

    Since R_D + R_B = 245, R_B = 245 - 108.9 = 136.1

    Books answer: R_B = 140N and R_C = 105N, have i done something wrong?






    2) A uniform plank AB has mass 40kg and length 4m. It is supported in a horizontal position by two smooth pivots. One pivot is at the end A and the other is at point C where AC = 3m. A man of mass 80kg stands on the plank which remains in equilibrium. The magnitude of the reaction at A is twice the magnitude of the reaction at C. The magnitude of the reaction at C is RN.

    Find the value of R.

    Right, so R_A = 2R_C
    => R_A - 2R_C = 0.................(1)
    Since its in equilibrium, R_A + R_C = 120g................(2)

    Solving as simultaneous equations, 3R_C = 120g, therefore R_C = 40g = 784.

    Taking clockwise moments about A: M(A) = 80g - (3 x 392) + (3 + x)80g = 0
    M(A) = 784 - 1176 + 2352 + 784x = 0
    784x = -1960
    x = -2.5m, however the answer is 0.5m, again, where have i gone wrong?

    all help is MUCH MUCH MUCH appreciated! thanks in advance
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    Draw a diagram for moments questions, it always helps

    For the first one, you are right to say that R_D + R_B = 245
    Then take the moments about B
    CLOCKWISE ... 2 x 245 = 490Nm
    ANTICLOCKWISE ... 3.5 x R_D
    Because it's in equilibrium 3.5 R_D = 490
    carry on from there....

    EDIT: Just realised where your mistake was on the first one. You said that the distance between B and D was 4.5 in your working, but in actual fact it is 3.5, (6 - 1 - 1.5 = 3.5) The rest of the method looks fine.
 
 
 
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