A 65 kg window cleaner uses a 14 kg ladder that is 4.2 m long. He places one end

Question

A 65 kg window cleaner uses a 14 kg ladder that is 4.2 m long. He places one end on the ground 1.5 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 3.4 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

Explanation / Answer

(a)

Balance torque about the lowermost point of the ladder,

14*9.8*(1.5/2) + 65*9.8*cos(theta)*1.5 = F*sqrt[(4.2)^2 - (1.5)^2]

where , F = Force on the window from the ladder

cos(theta) = 3.4 / 4.2

By solving,

F = 223.4 N

(b)

Force in horizontal direction,

F = Fx = 223.4 N

Force in vertical direction,

Fy = (65+14)*9.8 = 774.2 N

force on the ladder from the ground is,

F = sqrt (Fx^2 + Fy^2)

F = sqrt [(223.4)^2 + (774.2)^2]

F = 805.78 N

(c)

tan(theta) = Fy / Fx = 774.2 / 223.4

tan(theta) = 3.4655

theta = 73.9 deg

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.