A student pushes on a box weighing 500 N with a force F directed 30 degree below

Question

A student pushes on a box weighing 500 N with a force F directed 30 degree below the horizontal. What must F be such that the box starts to slide? If the student maintains the same force once the box starts to slide, what will be the acceleation? Assume the box and the floor are wood: use the value of the coefficient of ( static friction = 0.7 ).

Explanation / Answer

a) The normal force N = w + F sin      When box start to slide            F cos = f     ==> F cos = s N     ==> Fcos = sw + sFsin    ==> F = s w / cos -ssin              = 0.7 * 500 / cos30-0.7*sin30              = 678.28 N b)   a = [F cos - sw -sFsin ] / m           = [ 678.28cos30 - 0.7*500 - 0.7*678.28sin30] / 51.02           = 0.0001m/s2

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