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

2. The system in Fig. 2 is in equilibrium, with the string in the center exactly horizontal. Block X weighs 40 N, block Y weighs 50 N,
and angle φ is 35°.
Find (a) tension T1
(b) tension T2
(c) tension T3
(d) angle φ.

Fig 2.
Answer (a)  
ladder  
Since X and Y are 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.

Consider the forces at A.
A

T1h = T1 Sin 35°
T1v = T1 Cos 35
°

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

F can be resolved into F Sin θ and F Cos θ

Vertical forces: (using up as positive)
∑Vertical forces = 0

T1v + [-(W)] = 0
T1v = W
T1 Cos 35° = W
T1 = (40N) / Cos 35°
T1 = 49N

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 = -T1v


Horizontal forces: (using right as positive)
∑Horizontal forces = 0
T2L + [-(T1h)] = 0
T2L = T1h
T2L = T1 Sin 35° ---Eq1
T2L = (49N) Sin 35°
T2L = 28N

 

Consider the forces at B.
A

T3h = T3 Sin θ
T3v = T3 Cos θ

 
Vertical forces:
∑Vertical forces = 0
T3v + [-(W)] = 0
T3v = W
T3 Cos θ = W
T3 = (50N) / Cos θ ---Eq2

 

(b)
T3h + [-(T2R)] = 0
T3h = T2R
The tension on the horizontal string T2R = T2L
T3h = T2L
T3h = 28N
T3 Sin θ = 28N
T3 = 28N / Sin θ
From Eq2
50N / Cos θ = 28N / Sin θ
Sin θ / Cos θ = 28N / 50N
Tan θ = 28N / 50N
θ =Tan-1 (28N / 50N)
θ = 29°

T3 = 28N / Sin θ
T3 = 28N / Sin θ
T3 = 57N

Alternate Solution for T3.

using Pythagoras

T32 = T3v2 + T3h2

T32 = 502 + 282

T3 = 57N

 

 

 

 

 

Recall tan θ = Sin θ/Cos θ

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