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.

Explanation / Answer

minimum force required is Fmin = frictional force = F*cos(theta)

frictional force = mu_s*N

N is the normal force N+ F*sin(theta) = mg

N = mg- F*sin(30)

frictional force f = 0.405*(448-(F*0.5)) = 181.44 - 0.2025F = F*cos(30)

181.44-0.2025*F = 0.866*F

F = 169.8 N is the minimum force required

B)Since the table symmetrical

then Normal force on front legs is N/2 = (448-(169.8*0.5))/2 = 181.54 N

frictional force = f/2 = F*cos(30)/2 = (169.8*0.866)/2 = 73.5234 N

So back legs also experience the same frictional force and normal force

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