UNIT 4
Conditions for Static Equilibrium
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Mechanical Properties of Solids
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6

 

 

Questions and Answers on Static Equilibrium

ladder
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)  
ladder

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
Fig 3.1
 
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 (momentum)of its center of mass is zero.
— Its angular momentum (momentum_angular)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 AC2 = AB2 - BC2
AC2 = 52 - 2.52
AC = 4.3m

Vertical forces: (using up as positive)
∑Vertical forces = 0
[-(Wm)] + [-(Wl)] + Fg Sin θ= 0
(-750 N) + (-100N) + Fg Sin θ= 0
Fg Sin θ = 850N ---Eq1
Fg= 850N / Sin θ

resolvingRecall 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
Fw + [-(Fg Cos θ)]= 0
Fg Cos θ= Fw ---Eq2


Dividing Eq1 by Eq2
Tan θ = (850N) /(Fw)
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 vice-versa.


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
Fg2 = (Fg Sin θ)2 + (Fg Cos θ)2
Fg2 = 850 2 + 2882
Fg = 900N

   
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Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.