International Edition---engineering Mechanics: Statics, 4th Edition
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN: 9781305501607
Author: Andrew Pytel And Jaan Kiusalaas
Publisher: CENGAGE L
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Chapter 10, Problem 10.60P

The weightless bars AB and CE, together with the 5-lb weight BE, form a parallelogram linkage. The ideal spring attached to D has a free length of 2 in. and a stiffness of 7.5 lb/in. Find the two equilibrium positions that are in the range 0 < θ < π / 2 , and determine their stability. Neglect the weight of slider F.

Chapter 10, Problem 10.60P, The weightless bars AB and CE, together with the 5-lb weight BE, form a parallelogram linkage. The

Expert Solution & Answer
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To determine

The equilibrium positions.

Answer to Problem 10.60P

The system is at unstable equilibrium at 29.6o and at 53.1o the system comes in stable equilibrium.

Explanation of Solution

Given Information:

The weight of the load at BE = 5-lb

The stiffness of spring k = 7.5 lb/in

The following figure is given:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 10, Problem 10.60P , additional homework tip  1

Calculation:

Consider the following figure:

International Edition---engineering Mechanics: Statics, 4th Edition, Chapter 10, Problem 10.60P , additional homework tip  2

To calculate the angle for the equilibrium position, let us calculate the potential energy of the system. The total potential energy of the system consists of potential energy of the weight (Vg) and the potential energy of the springs (Ve)

The total potential energy = V=Vg+Ve

  V=(W×Yg)+(12×k×s2) ... (1)

  Yg= The vertical distance of center of gravity of weight above the reference lines = The elongation of the particular spring

  Here,Yg=10sinθ+2.5s=5cosθ

Putting the value of Yg and s in equation (1),

  V=(W×(10sinθ+2.5))+(12×k×( 5cosθ)2)

The potential energy of the system comes out to be

  V=(5×( 10sinθ+2.5))+(12×7.5× ( 5cosθ )2)V=50sinθ+12.5+93.75cos2θ ... (2)

Now, let us take the first derivative of the total potential energy of the system,

  dVdθ=50cosθ+2×93.75cosθ×(sinθ) ... (3)

  dVdθ=50cosθ187.5sinθcosθ

Now, the principle of minimum potential energy can be used to find the value of angle θ.

The roots of equation dVdθ=0 will provide the values of angles at equilibrium.

  dVdθ=050cosθ187.5sinθcosθ=0cosθ(50187.5sinθ)=0The roots of this equation arecosθ=0andsinθ=0.267The equlibrium positions arecosθ=0,θ=π2sinθ=0.267,θ=sin1(0.267),θ=29.6°andθ=53.1°

Now, differentiating equation 3 again to get d2Vdθ2

  d(sinθ.cosθ)dθ=cos2θsin2θ=cos2θ

  d2Vdθ2=50sinθ187.5(cos2θ)Nowforθ=π2d2Vdθ2=50sinπ2187.5(cos2×π2)( d 2 V d θ 2 )θ=π2=50+187.5( d 2 V d θ 2 )θ=π2=137.5(+ve)

Thus, system is at stable equilibrium at θ=π2.

  d2Vdθ2=50sinθ187.5(cos2θ)Nowforθ=29.6°d2Vdθ2=50sin29.6°187.5(cos2×29.6°)( d 2 V d θ 2 )θ=π2=24.6996( d 2 V d θ 2 )θ=π2=120.7(ve)

Thus, system is at unstable equilibrium at θ=29.6°.

  d2Vdθ2=50sinθ187.5(cos2θ)Nowforθ=53.1°d2Vdθ2=50sin53.1°187.5(cos2×53.1°)( d 2 V d θ 2 )θ=π2=39.98+52.31( d 2 V d θ 2 )θ=π2=12.33(+ve)

Thus, system is at stable equilibrium at θ=53.1°.

Conclusion:

Therefore, thesystem is at unstable equilibrium at 29.6o and at 53.1o the system comes in stable equilibrium.

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Chapter 10 Solutions

International Edition---engineering Mechanics: Statics, 4th Edition

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