UNIT 4
PHYS 2001
Example 01
Example 02
Example 03
Example 04
Example 05
Example 06
Example 07

 

 

Static Equilibrium - Example 06

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2. A leaf of mass 1 gram hangs on a rafter by a spider's thread as in fig 2b. The wind blows horizontally and the thread makes an angle of 20° with the vertical. Calculate (a) the tension on the thread and (b) the force of the wind on the leaf.

Fig 2a.
Answer  
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Forces on the Leaf [Fig 2 (i)]

Let T = Tension on the web
F = Force of the wind
W = Weight of the leaf

Weight of leaf (W) = mg
W = (1 X 10-3 kg) (10ms2)
W = (1 X 10-2 N)

Resolving T into vertical and horizontal components [Fig 3b(ii)]
Vertical Component = T Cos 20°
Horizontal component = T Sin 20°

Fig 2b.
 

(a) ∑F = 0

∑Vertical forces = 0
T Cos 20° + (-W) = 0
T Cos 20° = - (-W)
T Cos 20° = (1 X 10-2 N)
---Eq1
T = (1 X 10-2 N) / Cos 20°
T = (1.1 X 10-2 N)

∑Horizontal forces = 0
(-T Sin 20°) + F = 0
---Eq2
F = T Sin 20°
F = (3.6 X 10-3 N)

Double check Answer.
T Sin 20° = (1 X 10-2 kg) ---Eq1
T Cos 20° = - F ---Eq2 rearranged.
Dividing Eq1 by Eq 2
Tan 20° = (1 X 10-2 kg) / (- F)
(- F) = (1 X 10-2 kg) / Tan 20°
F = -(2.7 X 10-2 N)

ladderAlternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.

Cos 20° = W/T

T = W/Cos 20°

T = (0.01N)/Cos 20°

T = (1.1 X 10-2 N)

Tan 20° = F/W

F = Tan 20°/W

F = (Tan 20°)/(0.01N)

F = (3.6 X 10-3 N)

Double check:
(1.1 X 10-2 N)2 = (3.6 X 10-3 N)2 + (1.0 X 10-2 N)2

 
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3. A flower pot of mass 8kg is hung from two wires as show in in fig 3a. Find the tension in each wire.

Fig 3a.
Answer  
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Forces on the junction

T1 = Tension on the right wire
T2 = Tension on the left wire
W = Weight of the flower pot

Weight of flower pot (W) = mg
W = (8 kg) (10ms2)
W = (80 N)

 

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Resolving T1 into vertical and horizontal components
Vertical Component = T1 Sin 40°
Horizontal component = T1 Cos 40°

Resolving T2 into vertical and horizontal components
Vertical Component = T21 Sin 60°
Horizontal component = T2 Cos 60°

Fig 3b.
 

(a) ∑F = 0
∑Horizontal forces = 0
T1 Cos 40° + [-(T2 Cos 60°)] = 0
T1 = -[-(T2 Cos 60°) / (Cos 40°)]
T1 = 0.65 T2 ---Eq2

∑Vertical forces = 0
T1 Sin 40° + T2 Sin 60° + (-W) = 0
T1 Sin 40° + T2 Sin 60° = -(-80N)
T1 Sin 40° + T2 Sin 60° = (80N)
T1 Sin 40° = (80N) - T2 Sin 60°
T1 = {[(80N) / (Sin 40°) - (T2 Sin 60°)] / (Sin 40°)}
T1 = 124N - 1.35 T2 ---Eq1

Substuting for T1 from Eq1 into Eq2
0.65 T2 = 124N - 1.35 T2
124N = 1.35 T2 + 0.65 T2
124N = 2 T2
T2 = 62 N
T1 = 41N

ladderAlternative solution:
Since there is no resultant force, the three vectors must add up to zero, so they form a closed triangle.

Using (W/Sin 100° ) = (T1/Sin 30°) = (T2/Sin 50°)


(80N/Sin 100° ) = (T1/Sin 30°) = (T2/Sin 50°)

T1 = 80 Sin 30°/Sin 100°

T1 = 41N

T2 = 80 Sin 50°/Sin 100°

T2 = 62N

 

 

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