A symmetrical table of height 0.790 m, length 1.75 m, and weight 448 N is dragge

Question

A symmetrical table of height 0.790 m, length 1.75 m, and weight 448 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 0.405.

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

front legs: normal force

frictional force

back legs: normal force

frictional force

Explanation / Answer

(a) minimum force = umgSQRT(1+u^2) (here u = cofficient of friction )

so Fmin = 0.405*448SQRT(1+(0.405)^2)

Fmin = 181.44
.07 = 169.57N

(b) frictional force = 169.57*cos(30) = 36.66N frictional force is same for all legs

normal force = 169.57*sin(30) = 21.19N & normal force is also same for all legs

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