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13 August, 07:58

An initially stationary box of sand is to be pulled across a floor by means of a cable in which the tension should not exceed 1070 N. The coefficient of static friction between the box and the floor is 0.350. (a) What should be the angle between the cable and the horizontal in order to pull the greatest possible amount of sand, and (b) what is the weight of the sand and box in that situation?

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  1. 13 August, 11:57
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    a) The angle between the cable and the horizontal is 19°

    b) The weight of the sand and box = 3238.89N

    Explanation:

    Let T be the tension on the cable

    Let theta be the inclined angle at the horizontal

    Coefficient of static friction is 0.350

    Since box of sand is stationary, the net force is zero

    FN + Tsintheta - mg = 0

    Where FN is the magnitude of the normal force

    mg = weight of the box

    The normal force FN is given by:

    FN = mg - Tsin theta

    The horizontal component = Tcostheta-fs, max=0

    T = costheta - us (mg - Tsintheta) = 0

    T = (usmg) / (costheta + usintheta)

    a) Tan^-1 (us) = theta

    Tan^-1 (0.350) = 19°

    b) mg = T (cos19° + usin19°) / us

    mg = 1070 * (cos19 + 0.350sin19°) / 0.350

    mg = 1070 (0.9456+0.1139) / 0.350

    mg = 1070 (1.059) / 0.350

    mg = 3,238.89N
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