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
bartleby

Videos

Textbook Question
Book Icon
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
Check Mark
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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
The scissors linkage is subjected to a force of P = 150 N as shown in the figure below. Thespring is unstretched at θ = 0◦. All links have equal length of 0.3 m. Using the principles of virtualwork, determine the equation in terms of θ, for the system to be in equilibrium. Neglect the massof the links.    Find the numerical value of θ, by solving the equation that you have obtained,using a numerical method. Hint: You may look for a trial and error method, or Newton’s bisectionmethod, or Newton-Raphson method, etc.   Please solve the question
Cable AB passes over the small ideal pulley C without a change in its tension. What length of cable CD is required for static equilibrium in the position shown? What is the tension T in cable CD? 3' Answers: LCD = i TCD= i A 48° C B 59 lb ft lb D
The weight of the uniform bar AB is W. The stiffness of the ideal spring attached to B is k and the spring is unstreched when theta= 80 degree.if W=kL the bar has three equilibrium position in the range of 0

Chapter 10 Solutions

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

Ch. 10 - Determine the ratio P/Q of the forces that are...Ch. 10 - Find the vertical force P that will hold the...Ch. 10 - The linkage of the braking system consists of the...Ch. 10 - The automatic drilling robot must sustain a thrust...Ch. 10 - Determine the couple C for which the mechanism...Ch. 10 - The scissors jack is used to elevate the weight W....Ch. 10 - Prob. 10.17PCh. 10 - Calculate the torque C0 that must be applied to...Ch. 10 - Determine the force F and the angle a required to...Ch. 10 - Locate the instant center of rotation of bar AB...Ch. 10 - Prob. 10.21PCh. 10 - Determine the force P that will keep the mechanism...Ch. 10 - Prob. 10.23PCh. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Determine the ratio P/Q for which the linkage will...Ch. 10 - Prob. 10.27PCh. 10 - Prob. 10.28PCh. 10 - If the input force to the compound lever is P = 30...Ch. 10 - Determine the roller reaction at F due to the...Ch. 10 - Prob. 10.31PCh. 10 - Prob. 10.32PCh. 10 - Prob. 10.33PCh. 10 - Prob. 10.34PCh. 10 - Prob. 10.35PCh. 10 - For the pliers shown, determine the relationship...Ch. 10 - When activated by the force P, the gripper cm a...Ch. 10 - Prob. 10.38PCh. 10 - The hinge is of the type used on some automobiles,...Ch. 10 - The spring attached to the sliding collar is...Ch. 10 - The weight W is suspended from end B of the...Ch. 10 - The uniform bar of weight W and length L = 1.8R...Ch. 10 - A slender homogeneous bar is bent into a right...Ch. 10 - The body shown is a composite of a hemisphere and...Ch. 10 - Prob. 10.45PCh. 10 - The uniform bar AB of weight W and length L is...Ch. 10 - Uniform rods of weights W1 and W2 are welded to...Ch. 10 - Prob. 10.48PCh. 10 - The semi-cylinder of radius r is placed on a...Ch. 10 - Prob. 10.50PCh. 10 - The spring attached to the homogenous bar of...Ch. 10 - The spring is connected to a rope that passes over...Ch. 10 - Find the equilibrium positions of the 30-lb...Ch. 10 - The mechanism of negligible weight supports the...Ch. 10 - Solve Prob. 10.54 assuming that A and B are...Ch. 10 - The stiffness of the ideal spring that is...Ch. 10 - Find the stable equilibrium position of the system...Ch. 10 - The uniform bar AB of weight W = kL is in...Ch. 10 - The weight of the uniform bar AB is W. The...Ch. 10 - The weightless bars AB and CE, together with the...Ch. 10 - Prob. 10.61PCh. 10 - The bar ABC is supported by three identical, ideal...
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
International Edition---engineering Mechanics: St...
Mechanical Engineering
ISBN:9781305501607
Author:Andrew Pytel And Jaan Kiusalaas
Publisher:CENGAGE L
Mechanical SPRING DESIGN Strategy and Restrictions in Under 15 Minutes!; Author: Less Boring Lectures;https://www.youtube.com/watch?v=dsWQrzfQt3s;License: Standard Youtube License