A uniform ladder is 10.0-m long and weighs 200 N. The ladder leans against a ver

Question

A uniform ladder is 10.0-m long and weighs 200 N. The ladder leans against a vertical, frictionless wall at a height of 8.00 m above the ground. A horizontal force F~ a is applied to the ladder at a distance of 2.00 m from its base (measured along the ladder). The coefficent of static friction between the ladder and the ground is 0.500.

(a) What is the maximum value of F~ a that you can push on the ladder to the left without it beginning to slide towards the wall?

(b) What is the maximum value of F~ a that you can push on the ladder to the right without it beginning to slide away from the wall?

Explanation / Answer

angle made by the ladder with the ground=arcsin(8/10)=53.13 degrees

part a:

forces acting on the ladder:

let reaction force on the wall =F1 , to the right

normal force from the ground on the ladder=F2, in upward direction

friction force=friction coefficient*normal force=0.5*F2, to the right direction

weight of the ladder=200 N, in vertically downward direction

horizontal force =Fa , to the left, acting at a distance of 2 m from the base

height from the base of the ladder=2*sin(53.13)=1.6 m

balncing vertical forces,

F2=200 N

balancing horizontal forces:

Fa=F1+0.5*F2=F1+100...(1)

balancing torque about the base of the ladder:

F1*8=Fa*1.6+200*(6/2)=Fa*1.6+600

==>8*F1-1.6*Fa=600...(2)

solving equations 1 and 2 simultaneously,

F1=118.75 N

Fa=218.75 N

hence maximum force with which you can push the ladder to the left is 218.75 N

part b:

now Fa is directed to the right.

friction force of magnitude 0.5*F2 is directed to the left.

balancing forces in vertical direction:

F2=200 N

friction force=0.5*200=100 N

balancing forces in horizontal direction:

100=F1+Fa...(3)

balancing torque about base of the ladder:

F1*8+Fa*1.6=200*3=600...(4)

solving equations 3 and 4 simultaneously,

F1=68.75 N

Fa=31.25 N

so maximum force with which the ladder can be pushed to the right is 31.25 N

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