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

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4a. Find the tension in the two cords shown in Fig. 1. Neglect the mass of the cords, and assume that the angle is 30° and the mass Y is 100 kg.

Fig 1
Answer  

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Fig 2

(i) The force T1 can be resolved into two forces at right angles to each other T1h and T1v

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

so Sin 30° = T1v/T1 -----Eq 1

and Cos 30° = T1h/T1 --- Eq2

From Eq1 T1 = = T1h/Cos 30° --- Eq3

From Eq2 T1 = T1v/Sin 30° --- Eq4

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Fig 3

Figure 2 can be redone to show the components of Fg (Fv and Fh)

The system is in equilibrium:
Recall from Newton's First and Second Laws the conditions for static equilibrium:

∑F = 0
∑Horizontal forces = 0 --- Eq5
∑Vertical forces = 0
--- Eq6

 

From Eq6
T1v + [-(W)] = 0
T1v = W
Taking up as positive
T1v = (1000N)
T1v = 1000N up

From Eq4
T1 = = T1v/Sin 60°

T1 = 1000N/Sin 60°

T1 = 1155N


(iii) From Eq3
T1 = T1h/Cos 60°

T1 Cos 60° = T1h
T1h = 577N

From Eq5
T2 + [-(T1h)]= 0
T2 = T1h

Taking to the right as positive
T2 = (577N)
T2 = 577N

   
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4b. The crane begins to move until the rope makes an angle of 60° with the weight and the cable makes and angle θ with the rope. At this point the rope breaks at a tension of 3000N.
(i) Draw a diagram to represent the forces.
(ii) calculate the angle θ.
(iii) calculate the tension on the cable Tc.
Fig 4  
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(i) Tc = tension on cable

Tr = tension on rope = 3000N

W = Weight of the ball = 5000N


Fig 4
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In this example it is more convenient to resolves forces along the direction of the rope and perpendicular to the rope as the known and unknown angles are measured from the direction of the rope

Forces parallel to the direction of the rope:
Tr (to the left)
Tc Cos θ (Component of Tc to the right)
W Cos θ (Component of Tc to the left)

Forces perpendicular to the direction of the rope:
W (down)
W Sin 60° (Component of W down)
W Cos 60°(Component of W to the left)

 

The system is in equilibrium:
Recall from Newton's First and Second Laws the conditions for static equilibrium:

∑F = 0
∑ Forces parallel to direction of rope = 0 --- Eq1
∑Forces perpendicular to direction of rope = 0 --- Eq2

(ii) From Eq1:
[-(Tr)] + [-(W Cos 60°)]+ Tc Cos θ = 0
(-3000N) + (-5000 Cos 60°) + Tc Cos θ° =0
Tc Cos θ = -(-3000N) -(-5000 Cos 60°)

Tc = (3000N + 5000 Cos 60°) / Cos θ
Tc = 6500 /
Cos θ
Cos θ = 6500 / Tc ---Eq3

From Eq2:
[-(W Sin 60°)] + Tc Sin θ = 0
5000N Sin 60° = Tc Sin θ
Tc = -5000N Sin 60° / Sin θ
Tc = 4330 / Sin θ
Sin θ = 4330 / Tc ---Eq4

Dividing Eq4 by Eq 3
Sin θ / Cos θ = (4330 / Tc) / (6500/ Tc
Tan θ = 6500/4330
θ = tan-1 (4330/6500)
θ = 33.7°

(iii) From Eq3
Tc = 6500 / Cos θ
Tc = 7810N
From Eq4
Tc = 4330 / Sin θ
Tc = 7810N

   
Answers: 4 (a)
(i)

Double Checks for the answers Fg, Fv and Fh.
By Pythagoras:

T1h2 + T1hFv2 = T12

28872 + 50002 = 5773 2

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(ii) T1 = 5887N (to the right )
(iii) T2 = 2887N (to the left)
   
Answers: 4 (b)

 

 

 

(i)

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(ii) θ = 33.7°
(iii) Tc = 7810N

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