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Question

A uniform ladder 7.0 m long weighing 450N rests with one end on the ground and the other end against a perfectly smooth vertical wall.The ladder rises at 60 degree above the horizontal floor. a 750 newton painter finds that she can climb 2.75m up the ladder. a) what fall does the wall exert on the ladder? b)find the friction force and the normal force that the floor exerts on the ladder.

Answer 1

What do you mean? A ladder can't be 450N, N=Newtons which is a measurement of force

Answer 2

a. the force exerted on the ladder by the wall is equal in magnitude to the frictional force acting on the ladder.

Fp = weight of person on ladder = 750N

Fl = weight of the ladder = 450N

Fw = reaction from wall

Fn = reaction from ground

Ff = frictional force

Equilibrium of forces:

Fw = Ff .................. horizontal

Fn = Fp + Fl ........... vertical

Equilibrium of moments about the base of the ladder:

phi = angle ladder makes with wall = 90 - 60 = 30 degrees

s = distance up the ladder of the painter = 2.7 m

c = length of ladder = 7 m

Fw*c*cos(phi) = Fp*s*sin(phi) + Fl*(c/2)*sin(phi)

Fw(7)(SQRT(3)/2) = (750)(2.7)(1/2) + (450)(7/2)(1/2)

Fw(6.0622) = 945 + 787.5 = 1732.5

Fw =  285.787 = Ff

The wall exerts a force of 285.787 N on the ladder and this is the same as the frictional force although in opposite directions

Fn = Fp + Fl = 450 + 750 = 1200

The normal force from the floor is 1200 N

The ladder is about to slipe when Ff = mu*Fn

285.787 = mu*(1200)

mu = 0.238 is the coefficient of friction