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. 164.45 N
  

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

front legs:	normal force
	frictional force
back legs:	normal force
	frictional force

Answer 1

a)

let the force needed is F

in vertical direction

N is the normal force

462 - F * sin(30) = N

for the force in horizontal direction

F * cos(30) = u * N

F * cos(30) = 0.375 * (462 - F * sin(30))

solving for F

F = 164.4 N

the force needed to move the table is 164.4 N

part B)

for the front legs ,

balancing the moment of forces about back legs

Nf * 1.55 - 462 * (1.55/2) - 164.4 * cos(30) * 0.75 + 164.4 * 1.55 * sin(30) = 0

solving for Nf

Nf = 217.7 N

the normal force at the front legs is 217.7 N

frictional force = u * N

frictional force = 0.375 * 217.7

frictional force = 81.6 N

the frictional force at the front legs is 81.6 N

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Now , for the back legs

balancing the forces in vertical directio

Nb + 164.4 * sin(30) + 217.7 - 462 = 0

Nb = 162.1 N

the normal force at the back legs is 162.1 N

frictional force = u * Nb

frictional force = 0.375 * 162.1

frictional force = 60.8 N

the frictional force at the back legs is 60.8 N