

Questions and Answers on Static Equilibrium

3. A 75 kg window cleaner uses a 10 kg ladder that is 5.0 m long.
He places one end on the ground 2.5 m from a wall, rests the upper
end against a cracked window, and climbs the ladder. He is 3.0 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? 
Fig 3.0 

Answer
(a) 


Wl = 10kg X 10ms^{2} = 100N
Wm = 75kg
X 10ms^{2} = 750N
AB = Length of ladder = 5.0m
BC = Distance of ladder foot from the wall = 2.5m
Fw = Reaction from Wall
Fg = Reaction from floor

5.0m
Since the ladder is in equilibrium:
∑F = 0
∑Horizontal forces = 0
∑Vertical forces = 0
∑Torques about any point = 0 
Supplementary Explanations:
Recall from Newton's First and Second Laws the conditions for static equilibrium:
— The linear momentum ()of its center of mass is zero.
— Its angular momentum ()about its center of mass, or about any other point, is also zero. 
(a) Taking moments about B (anticlockwise is positive)
Fg X (0) + Wm X (½BC) + Wl X (½BC) + [(Fw X (AC))] = 0
(0) + Wm X (1.2m) + Wl X (½BC) + [(Fw X (1.5m))] = 0
(750N) X (1.2m) + (100N) X (1.5m) + [(Fw X (4.3m))] = 0
Fw = [(750N X 1.5) + (100N X 1.25)] / (4.3)
Fw = 288N 
Cos ∠ABC = BC / AB
BD = 3m Cos ∠ABC
BD = {[3 (BC)] / (AB)}
BD = 1.5 m
OR Alternatively
Triangles ABC and BDE are similar so BD/BC = BE/AB
BD = (BE) (BC) / AB
BD = (3m) (2.5m) / (5m)
BD = 1.5m
By Pythagoras AC^{2} = AB^{2}  BC^{2}
AC^{2} = 5^{2}  2.5^{2}
AC = 4.3m

Vertical forces: (using up as positive)
∑Vertical forces = 0
[(W_{m})] + [(W_{l})] + Fg Sin θ= 0
(750
N) + (100N) + Fg Sin θ= 0
Fg Sin θ = 850N Eq1
Fg= 850N / Sin θ

Recall Sin θ = ^{opposite}/_{hypotenuse}
and Cos θ = ^{adjacent}/_{hypotenuse}
Fg can be resolved into Fg Sin θ and Fg Cos θ 
(c) Horizontal forces: (using right as positive)
∑Horizontal forces = 0
F_{w} + [(Fg Cos θ)]= 0
Fg Cos θ= F_{w} Eq2
Dividing Eq1 by Eq2
Tan θ = (850N) /(F_{w})
Tan θ = (850N) /(288N)
θ = Tan^{1} (850N) /(288N)
θ = 71°

Recall that we can treat with horizontal and vertical forces independently as a force has no component at right angles to it so vertical forces can have no horizontal forces and viceversa.
Recall Tan θ = ^{Cos θ}/_{Sin θ}

(c)
Fg Sin θ = 850N Eq1
Fg= 850N / Sin θ
Fg = 900N

Alternate Solution for Fg _{}.
using Pythagoras
Fg_{}^{2} = (Fg Sin θ)^{2} + (Fg Cos θ)^{2}
Fg_{}^{2} = 850 ^{2} _{} + 288^{2}
Fg_{} = 900N 

