A painter is standing on the ladder (mass 40 kg and length 2.5 m) in the figure

Question

A painter is standing on the ladder (mass 40 kg and length 2.5 m) in the figure below. There is friction between the bottom of the ladder and the floor with

?S = 0.30,

but there is no friction between the ladder and the wall. The painter has mass 70 kg and is standing a distance of 0.60 m from the top.

(a) Make a sketch showing all the forces on the ladder. Be sure to show where each force acts.

Score: 0.8 out of 0.8

Comment:

(b) Express the conditions for translational and rotational equilibrium for the ladder. (Use the following as necessary: Ff for frictional force between the ladder and floor,mP for the mass of the painter, mL for the mass of the ladder, LP for the painter's distance from the ladder's bottom, L for the length of the ladder, NF for the normal force where the ladder meets the floor, NW for the normal force where the ladder meets the wall, and ? for the angle formed between the ladder and the floor. Assume +x direction is toward the right and the +y direction is upward. )

Fx = ----------? = 0
Fy = ----------? = 0
? = -----------? = 0
(c) If

? = 64

Explanation / Answer

Fx = Ff - Nw = 0

Fy = Nf - (m_l + m_p)*g = 0

Torque T = Nw*(L*sin theta) - (m_l*g)*((L*cos theta)/2) - (m_p*g)*(L_p*cos theta) = 0

c)

Nf = (40+70)*9.81 = 1079.1 Newton

Ff = 0.3*Nf = 323.73 N

Nw = Ff = 323.73 N

Torque = 323.73*2.5*sin64 - 40*9.81*(2.5/2)*cos64 - (70*9.81)*(2.5 - 0.6)*cos64

= -59.56 Nm (minus sign denotes anticlockwise)

Since direction = anticlockwise, ladder will not slip.

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