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sir lancelot, who weighs 800 n, is assaulting a castle by climbing a uniform ladder that is 5.0 m long and weighs 180 n. the bottom of the ladder rests on a ledge and leans across the moat in equilibrium against a frictionless, vertical castle wall. the ladder makes an angle of 53.1 deg with the horizontal. lancelot pauses one third of the way up the ladder. (a) find the normal force and friction forces on the base of the ladder. (b) find the minimum coefficient of static friction needed to prevent slipping at the base. (c) find the magnitude and direction of the contact force on the base of the ladder. (that is, the vector sum of the forces at the base.)



Answer :

(a) The normal force is 267.7 N and frictional force is 980 N.

(b) The minimum coefficient of static friction is 0.27, which is needed to prevent slipping at the base.

(c) The direction of the contact force is 1016 N.

∈(t) = 0

N₂ ( L sinθ ) = 800 ( [tex]\frac{L}{3}[/tex] cosθ )

N₂ = 267.7 N

(a) Normal force F(d) = N₂ = 267.7 N

Frictional force N = 800 + 180

                         N = 980 N

(b) The minimum coefficient of static friction,

F(d) = μ N₁

μ = F / N₁

μ = 267.7 / 980

μ = 0.27

(c) The direction of the contact force,

R = [tex]\sqrt{(980)^{2} + (267.7)^{2} }[/tex]

R = 1016 N

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