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

1.A uniform ladder of length 10m and mass 10kg leans against a wall as in Fig 1 at an angle of 60°.
(a) Consider there is no friction between the wall and the top of the ladder, draw a diagram to represent the forces on the ladder.
(b) Calculate the resultant force from the ground and its angle, θ, to the horizontal.
(c) Calculate the friction force on the ladder from the floor.
(d) Calculate the normal reaction from the floor.

Fig1
Answer  

(a)

ladder

 
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.
Fg can be resolved into its horizontal and vertical components.
Fghorizontal = Fg Cos θ
Fgvertical = Fg Sin θ

resolvingRecall Sin θ = opposite/hypotenuse
and Cos θ = adjacent/hypotenuse

Fghorizontal is actually the friction from the ground

Fgvertical is the normal reaction of the floor

Vertical forces:
[-(W)] + Fgvertical = 0 ---Eq1
[-(W)] + Fg Sin θ = 0
Fg Sin θ = W
Sin θ = W/Fg
Sin θ = 100/Fg --- Eq2
Sin2θ = [(100/Fg)]2 --- Eq3

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.

W = -Fgvertical
Fgvertical is the normal reaction of the floor

Horizontal forces:
Fw + [-Fghorizontal)] = 0 ---Eq4
Fw + [-(Fg Cos θ)] = 0
Fg Cos θ = Fw
Cos θ = Fw/Fg --- Eq5
Cos2θ = [(Fw/Fg)2] --- Eq6
Fw = -Fghorizontal
Fghorizontal is the friction from the floor
Taking moments about B (anticlockwise is positive)
Fg X (0) + [-(W X (½BC))] + [-(Fw X (AC))] = 0
[-(100N X (½ X 10 Cos 60°)] + [-(Fw X 10 Sin 60°)] = 0
500 Cos 60° = 10 Fw Sin 60°
Fw Sin 60° = 50 Cos 60°
Fw = (50 Cos 60°)/(Sin 60°)
Fw = 50/(tan 60°)
Fw = 28.9N

Recall moment of a force = Force X perpendicular distance to force
Clockwise moments = Anticlockwise moments when in equilibrium.

We take moments through the point which Fg acts so its moment is zero.

Recall tan θ = Sin θ/Cos θ

 

(b) Solving Fg and θ
Adding Eq3 and Eq6

Cos2θ + Sin2θ = [(100/Fg)]2 + [(Fw/Fg)2]
1 = [(100/Fg)]2 + [(Fw/Fg)2]
1 = 10000 / Fg2 + Fw2 / Fg2
1 = (10000 + Fw2) / Fg2
Fg2 = 10000 + Fw2
Fg2 = 10000 + 28.92
Fg = 10000 + 28.92
Fg = 104N

From Eq2:
θ = Sin-1(100/104)
From Eq5:
θ = Cos -1(28.9/104)

θ = 74°

Recall Cos2θ + Sin2θ = 1

Double Check:
Fg2 = Fgvertical2 + Fghorizontal2
1042 = 28.92 + 1002

Also θ = tan -1(100/28.9)

(c) From Eq4:
Fghorizontal = Fw

Friction = 28.9N

Double Check:
Fghorizontal = Fg Cos θ
Fghorizontal = 104 Cos 74°

(d) From Eq1:
Fgvertical = W
Normal Reaction from floor = 100N
Double Check:
Fgvertical = Fg Sin θ
Fghorizontal = 104 Sin 74°
   
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Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.