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 10.97P

Bars AB and BC, each with a length l and of negligible weight, are attached to two springs, each of constant k. The springs are 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 10.97P, Bars AB and BC, each with a length l and of negligible weight, are attached to two springs, each of

Fig. P10.97

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 10.97P

The range of values of P for which the equilibrium position is stable is 0P<0.382kl_.

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 10.97P

Find the horizontal distance (xB) using the relation.

xB=lsinθ1

Find the horizontal distance (xC) using the relation.

xC=lsinθ1+lsinθ2

Find the vertical distance (yC) using the relation.

yC=lcosθ1+lcosθ2

When the values are small,

sinθ1θ1;sinθ2θ2cosθ11θ122;cosθ21θ222

Find the potential energy (V) using the relation.

V=Vg+Vs=PyC+12kxB2+12kxC2

Here, the magnitude of the force applied at C is P and the spring constant is k.

Substitute (lcosθ1+lcosθ2) for yC, lsinθ1 for xB, and (lsinθ1+lsinθ2) for xC.

V=P(lcosθ1+lcosθ2)+12k(lsinθ1)2+12k(lsinθ1+lsinθ2)2

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

V=P(l(1θ122)+l(1θ222))+12k(lθ1)2+12k(lθ1+lθ2)2=Pl(1θ122+1θ222)+12kl2(θ12+(θ1+θ2)2) (1)

Differentiate the Equation (1) with respect to θ1.

Vθ1=Pl(2θ12)+12kl2(2θ1+2(θ1+θ2))=Plθ1+kl2(2θ1+θ2) (2)

Differentiate the Equation (2).

2Vθ12=Pl+2kl2

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

2Vθ1θ2=kl2

Differentiate the Equation (1) with respect to θ2.

Vθ2=Pl(2×θ22)+12kl2(2(θ1+θ2))=Plθ2+kl2(θ1+θ2) (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=Pl(0)+kl2(2(0)+(0))=0

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

Vθ2=Pl(0)+kl2(0+0)=0

Vθ1=Vθ2=0

The condition is satisfied. The equilibrium is stable.

Condition 2:

Check the condition,

(2Vθ1θ2)22Vθ122Vθ22<0

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

(kl2)2(Pl+2kl2)(Pl+kl2)<0k2l4P2l2+Pkl3+2Pkl32k2l4<0P2+3Pklk2l2<0P23Pkl+k2l2>0

Solve the equation using the mathematical equation.

P<352kl<0.382kl

Condition 3;

Check the condition;

2Vθ12>0Pl+2kl2>0P+2kl>0P<2kl

Condition 4:

2Vθ22>0Pl+kl2>0P+kl>0P<kl

Refer to all the conditions,

The minimum value of P is 0.

The maximum value of P is 0.382kl.

Therefore, the range of values of P for which the equilibrium position is stable is 0P<0.382kl_.

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Chapter 10 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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