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

5. A planter of weight W1, 500N, is suspended from the end of a (hinged) uniform strut of length L, 2.00 m, and weight W2, 100N, as shown. The cable connecting the beam to the wall is horizontal.
(a) Find the tension in the cable and the reaction force exerted on the strut by the hinge.
(b) If the cable is made of a material with Yield Strength 3.4 x 108 Nm-2, what is the minimum radius of the horizontal cable to prevent permanently stretch of the cable?

Fig 5.
ladder

T = Tension in the cable
FH = Force exerted on the strut by the hinge

FHV = FH Sin θ

FHH = FH Cos θ


WZ2 = Weight of planter
W2 = 500N
W1 = Weight of strut
W1 = 100N

BE = 2.00m Cos 30°
BE = 1.73m
FE = 2.00m Sin 30°
FE = 1.00m

Answer  

(a) The strut is in equilibrium so
∑Torques about any point = 0
Clockwise Torques + Anticlockwise Torques = 0.

Taking Torgues at B
τclockwise+ τanticlockwise = 0

[-(W1 X BE)] + [-(W2 X ½BE)] + (T X FE)] = 0
[-(500N X 1.73m)] + [-(100N X 0.5m)] + (T X 1m)] = 0

- (866) -(50) + T = 0

T = 916N


We choose to take moments about B as it is difficult to fing the torque of FH as we don't know the angle.
Supplementary Explanations:
Recall from Newton's First and Second Laws the conditions for static equilibrium:
— Its angular momentum (momentum_angular)about its center of mass, or about any other point, is also zero.

∑F = 0

∑Horizontal forces = 0
FHH + [-(T)] = 0
FHH + [-(916N)] = 0
FHH = 916N

∑Vertical forces = 0
FHV + [-(W1)] + [-(W2)] = 0
FHV + [-(100N)] + [-(500N)] = 0
FHV = 600N

Now FHH = FH Cos θ
and FHV = FH Sin θ
So tan θ = [(FHV)/ (FHH)]
θ = 33.2°

FH = FHV / Sin θ
FH = 600 / Sin 33.2°
FH = 1100N

FH = FHH / Cos θ
FH = 916N / Cos 33.2°
FH = 1110N


Supplementary Explanations:
Recall from Newton's First and Second Laws the conditions for static equilibrium:
— Its angular momentum (momentum_angular)about its center of mass, or about any other point, is also zero.

The yield strength is the minimum stress which produces permanent plastic deformation.

Stress = F/A
3.4 x 108 Nm-2 = (1110 N) / (A)
(A) = (1110 N) / 3.4 x 108 Nm-2
(πr2) = 3.26 x 10-6 m2
r = 1.81 X 10-3 m
r = 1.81 mm.

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