a 5.00-m-long ladder, weighing 200 n, rests against a smooth vertical wall with its base on a horizontal rough floor, a distance of 1.20 m away from the wall. the center of mass of the ladder is 2.50 m from its base, and the coefficient of static friction between the ladder and the floor is 0.200. how far up the ladder, measured along the ladder, can a 600-n person climb before the ladder begins to slip?

When a 5.00 m-long ladder, with all weighing of 200 n, rests among a smooth vertical type of wall, then the ladder, will be measured among the the height = 0.488 m.

If θ be the angle ladder makes with the plane

cos θ = 1.2 / 5

Tan θ = 4.04

Let the height a person of weight 600 N can climb be h from the ground .

Distance from the base point where ladder touches the floor = h / tanθ

= h / 4.04

Total reaction force = total downward force

R = 200 + 600

800 N

Frictional force = μ R

= .2 x 800

= 160 N

Taking moment of force about the point on the ladder where it touches the floor and balancing them

= 200 x 1.2 x .5 + 600 x h / tanθ = μ R x 1.2 / tanθ ( reaction at the top point of ladder where it touches the wall is R₁ and R₁ =μ R )

= 200 x 1.2 x .5 + 600 x h / tanθ = 160 x 1.2 / tanθ

120 - 600 h / 4.04 = 47.52

120 - 47.52 = 600 h / 4.04

72.48= 148.51 h

h = 0.488 m

To learn more about friction, refer: brainly.com/question/24338873

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