A symmetrical table of height 0.750 m, length 1.55 m, and weight 462 N is dragge

Question

A symmetrical table of height 0.750 m, length 1.55 m, and weight 462 N is dragged across the floor by a force applied to its front edge. The force is directed to the right and upward and makes an angle of 30.0° with the horizontal.

(a) Find the minimum force necessary to drag the table across the floor. The coefficient of sliding friction between table and floor is 0.375.


(b) Calculate the normal force and the frictional force on each leg.

back legs:

back legs:

Explanation / Answer

normal force   N = m*g - F*sin30

frictional force = f =u*N

f = u*(mg - F*sin30)

along horizantal f = Fx = F*cos30

F*cos30 = u*(mg - F*sin30)

F*cos30 = 0.375*((462) - (F*sin30))

F = 164.45 N

b)

Normal force = (m*g - F*sin30)/2 = (462-(164.45*sin30))/2 = 190 N

frictional force = u*N = 71.25 N

Normal force = (m*g - F*sin30)/2 = (462-(164.45*sin30))/2 = 190 N

frictional force = u*N = 71.25 N

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