Vector Mechanics for Engineers: Statics
Vector Mechanics for Engineers: Statics
12th Edition
ISBN: 9781259977268
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek
Publisher: McGraw-Hill Education
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Chapter 10.2, Problem 10.80P

A slender rod AB with a weight W is attached to two blocks A and B that can move freely in the guides shown. Knowing that the spring is unstretched when AB is horizontal, determine three values of θ corresponding to equilibrium when W = 300 lb, l = 16 in., and k = 75 lb/in. State in each case whether the equilibrium is stable, unstable, or neutral.

Chapter 10.2, Problem 10.80P, A slender rod AB with a weight W is attached totwo blocks A and B that can move freely in theguides

Fig. P10.79 and P10.80

Expert Solution & Answer
Check Mark
To determine

Find three values of θ corresponding to equilibrium and check whether the equilibrium is stable, unstable, or neutral.

Answer to Problem 10.80P

The value of θ for equilibrium is θ=9.39°;θ=34.2°;θ=90°_.

The equilibrium is stableforθ=9.39°_.

The equilibrium is unstableforθ=34.2°_.

The equilibrium is stableforθ=90°_.

Explanation of Solution

Given information:

The weight of the slender rod AB is W=300lb.

The length of the slender rod AB is l=16in..

The spring constant is k=75lb/in..

Calculation:

The spring is unstretched when AB is horizontal.

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

Vector Mechanics for Engineers: Statics, Chapter 10.2, Problem 10.80P

Find the elongation of the spring s using the relation.

s=(lsinθ+lcosθ)l=l(sinθ+cosθ1)

Here, the length of the slender rod AB is l.

Find the potential energy (V) using the relation;

V=Vs+Vg=12ks2+W(l2sinθ)=12k(l(sinθ+cosθ1))2Wl2sinθ=12kl2(sinθ+cosθ1)2Wl2sinθ

Differentiate the equation;

dVdθ=kl2(sinθ+cosθ1)(cosθsinθ)Wl2cosθ (1)

Here, the weight of the slender rod AB and the spring constant is k.

Differentiate the Equation (1).

d2Vdθ2=kl2[(cosθsinθ)(cosθsinθ)+(sinθ+cosθ1)(sinθcosθ)]+Wl2sinθ=kl2[cos2θsinθcosθsinθcosθ+sin2θsin2θsinθcosθ+sinθsinθcosθcos2θ+cosθ]+Wl2sinθ=kl2[4sinθcosθ+sinθ+cosθ]+Wl2sinθ=kl2[2sin2θ+sinθ+cosθ]+Wl2sinθ (2)

Equate the Equation (1) to zero for equilibrium.

dVdθ=0kl2(sinθ+cosθ1)(cosθsinθ)Wl2cosθ=0cosθ[kl(sinθ+cosθ1)(1tanθ)W2]=0kl(sinθ+cosθ1)(1tanθ)=W2;cosθ=0

Substitute 300 lb for W, 75lb/in. for k, and 16 in. for l.

75×16×(sinθ+cosθ1)(1tanθ)=3002(sinθ+cosθ1)(1tanθ)=0.125

Solve the equation by trial and error procedure.

θ=9.39°;θ=34.2°;forkl(sinθ+cosθ1)(1tanθ)=W2θ=90°;forcosθ=0

Therefore, the value of θ for equilibrium is θ=9.39°;θ=34.2°;θ=90°_.

For θ=9.39°;

Substitute 300 lb for W, 75lb/in. for k, 16 in. for l, and 9.39° for θ in Equation (2).

d2Vdθ2=75×162×[2sin2(9.39°)+sin9.39°+cos9.39°]+300×162sin9.39°=19,200[2sin2(9.39°)+sin9.39°+cos9.39°]+2,400sin9.39°=10,104.54lbin.>0,stable

Therefore, the equilibrium is stableforθ=9.39°_.

For θ=34.2°;

Substitute 300 lb for W, 75lb/in. for k, 16 in. for l, and 34.2° for θ in Equation (2).

d2Vdθ2=75×162×[2sin2(34.2°)+sin34.2°+cos34.2°]+300×162sin34.2°=19,200[2sin2(34.2°)+sin34.2°+cos34.2°]+2,400sin34.2°=7,682.5lbin.<0,unstable

Therefore, the equilibrium is unstableforθ=34.2°_.

For θ=90°;

Substitute 300 lb for W, 75lb/in. for k, 16 in. for l, and 90° for θ in Equation (2).

d2Vdθ2=75×162×[2sin2(90°)+sin90°+cos90°]+300×162sin90°=19,200[2sin2(90°)+sin90°+cos90°]+2,400sin90°=21,600lbin.>0,stable

Therefore, the equilibrium is stableforθ=90°_.

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Chapter 10 Solutions

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