A painter is standing on a ladder (mass 40 kg.and length 2.5 m). There is fricti

Question

A painter is standing on a ladder (mass 40 kg.and length 2.5 m). There is friction between the bottom of the ladder and the floor with Us=0.30, but there is no friction between the ladder and the wall. The painter of mass 70kg. is standing a distance of 0.6 m. from the top of the ladder.
a) Express the conditions for translational and rotational equilibrium for the ladder?
b) If theta = 60 degrees (angle with bottom of ladder), use the results from part a and determine wether or not this ladder will slip?

Explanation / Answer

let's call the forces:

Nv=force of floor acting up
f=frictional force of floor (acting to left since I assume the wall is to the left); this force is also written as uNv where u is the coefficient of friction and Nv is the normal force acting on the ladder/painter

Fh=reaction force of wall acting horizontally to the right
Wl=weight of the ladder = 40g=392N
Wp=weight of the painter =70g=686N

horizontal equilibrium implies:

Fh=friction = u Nv

vertical equilibrium implies:

Nv=Wl+Wp=1078 N
therefore, Fh=uNv=0.3x1078=323.4N

for rotational equilibrium, the torques will sum to zero around any point we choose as the pivot; for simplicity, let's choose the point where the ladder touches the floor

rotational equilibrium: sum of torques =0

there is no torque due to Nv and friction at the chosen point since their action lines pass through the point

torque due to Wl= weight of ladder x 1/2 length ladder x cos(theta) =(2.5/2) (392)cos(theta) where theta is the angle between the ladder and floor; this produes a counterclockwise torque

torque due to Wl=490 cos(theta)

torque due to painter = Wp x 1.9 cos(theta)
torque due to Wp=1303.4 cos (theta) counterclocwise

torque due to Fh (2.5) sin(theta) =323.4(2.5 sin(theta)) =
808.5 sin(theta) clockwise

so we have from rotational equilibrium:

808.5sin(theta)=cos(theta)[1303.4+490]
808.5 sin(theta)=cos(theta)(1793.4)
tan(theta)=2.18=>theta=65.7deg

this is the minimum angle for which equilibrium is possible

at 60 degrees, the ladder will slip

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