A symmetrical table of height 0.860 m, length 1.40 m, and weight 460 N is dragge

Question

A symmetrical table of height 0.860 m, length 1.40 m, and weight 460 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.

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

Explanation / Answer

Here ,

a) let the foece applied is F

for the normal force N

N = W - F * sin(30)

N = 460 - F * sin(30)

in horizontal direction , for moving the table

F * cos(30) = frictional force

F * cos(30) = 0.375 * ( 460 - F * sin(30))

solving for F

F = 163.7 N

the minimum force needed to drag the table is 163.7 N

b)

for the normal force on front legs , Nf

balancing the moment of force about the back legs

460 * 1.4/2 - Nf * 1.4 - 163.7 * sin(30)* 1.4 + 163.7 * cos(30) * 0.86 = 0

solving for Nf

Nf = 235.2 N

the noral force at the front legs is 235.2 N

frictional force at the front legs = 0.375 * 235.2

frictional force at the front legs = 88.2 N

-------------------------

at the back legs

balancing the force in vertical direction

Nb + Nf + F * sin(30) = 430

Nb + 235.2 + 163.7 * sin(30) = 430

Nb = 113 N

the Nornal force at the back legs is 113 N

frictional force = 0.375 * 113

frictional force = 42.4 N

the frictional force at the back legs is 42.4 N

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