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Figure 6-24 shows an initially stationary block of mass m on a floor. A force of magnitude 0.540mg is then applied at upward angle 16 degrees. What is the magnitude of the acceleration of the block across the floor if (a)Us=.6 and Uk=.5 and (b) Us=.4 and Uk=.3?

  • Physics -

    Is the 0.540mg 0.540 milligrams or 0.54m * g? I will assume the latter.

    For case (a), the maximum possible static friction force is
    Fs = (m*g cos16* - 0.54m*g sin16)*Us
    = 0.487 m g
    The horizontal component of applied force is 0.540m*g*cos 16 = 0.519 m*g. Since this exceeds the maximum possible static friction force, the block slips and accelerates at a rate determined by
    kinetic friction.

    The friction force in the horizontal direction must be recomputed and subtracted from the applied force component in that direction.

    For case (b), proceed similarly

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