Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 10.2, Problem 10.100P

Solve Prob. 10.99, assuming that the vertical force applied at B is increased to 5P.

*10.99 Bars AB and CD, each of length l and of negligible weight, are attached to a spring of constant k. The spring is undeformed and the system is in equilibrium when θ1 = θ2 = 0. Determine the range of values of P for which the equilibrium position is stable.

Chapter 10.2, Problem 10.100P, Solve Prob. 10.99, assuming that the vertical force applied at B is increased to 5P. 10.99 Bars AB

Fig. P10.99

Expert Solution & Answer
Check Mark
To determine

Find the range of values of P for which the equilibrium of the system is stable.

Answer to Problem 10.100P

The range of values of P for which the equilibrium position is stable is P<4kl5_.

Explanation of Solution

Given information:

The system is in equilibrium, when θ1=θ2=0.

Calculation:

Show the free-body diagram of the arrangement as in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 10.2, Problem 10.100P

Find the horizontal distance (xB) using the relation.

xB=lsinθ1

Find the horizontal distance (xC) using the relation.

xC=lsinθ2

Find the vertical distance (yB) using the relation.

yB=lcosθ1

Find the vertical distance (yC) using the relation.

yC=lcosθ2

When the values are small,

sinθ1θ1;sinθ2θ2cosθ11θ122;cosθ21θ222

Find the potential energy (V) using the relation.

V=Vg+Vs=5PyB+PyC+12k(xCxB)2

Here, the spring constant is k.

Substitute lcosθ1 for yB, lcosθ2 for yC, lsinθ2 for xC, and lsinθ1 for xB.

V=5P(lcosθ1)+P(lcosθ2)+12k(lsinθ2lsinθ1)2=5Plcosθ1+Plcosθ2+kl22(sinθ2sinθ1)2

Substitute θ1 for sinθ1, θ2 for sinθ2, (1θ122) for cosθ1, and (1θ222) for cosθ2.

V=5Pl(1θ122)+Pl(1θ222)+kl22(θ2θ1)2 (1)

Differentiate the Equation (1) with respect to θ1.

Vθ1=5Pl(2θ12)+Pl(0)2kl22(θ2θ1)=5Plθ1kl2(θ2θ1) (2)

Differentiate the Equation (2).

2Vθ12=5Pl+kl2

Differentiate the equation (2) with θ2 to find the derivative of 2Vθ1θ2.

2Vθ1θ2=kl2

Differentiate the Equation (1) with respect to θ2.

Vθ2=5Pl(0)+Pl(2θ22)+2kl22(θ2θ1)=Plθ2+kl2(θ2θ1) (3)

Differentiate the Equation;

2Vθ22=Pl+kl2

Condition 1:

When the equilibrium is stable, θ1=θ2=0.

Substitute 0 for θ1 and 0 for θ2 in Equation (2).

Vθ1=5Pl(0)kl2(00)=0

Substitute 0 for θ1 and 0 for θ2 in Equation (3).

Vθ2=Pl(0)+kl2(00)=0

Vθ1=Vθ2=0

The condition is satisfied. The equilibrium is stable.

Condition 2:

Check the condition,

(2Vθ1θ2)22Vθ122Vθ22<0

Substitute kl2 for 2Vθ1θ2, (2Pl+kl2) for 2Vθ12, and (Pl+kl2) for 2Vθ22.

(kl2)2(5Pl+kl2)(Pl+kl2)<0k2l4+5P2l25Pkl3+Pkl3k2l4<05P4kl<0P<4kl5

Condition 3:

2Vθ22>0Pl+kl2>0Pl<kl2P<kl

Refer to all the conditions,

The minimum value of P is 0.

The maximum value of P is Pmax<4kl5.

Therefore, the range of values of P for which the equilibrium position is stable is P<4kl5_.

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Chapter 10 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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