solution

Published on February 2017 | Categories: Documents | Downloads: 33 | Comments: 0 | Views: 174
of 1
Download PDF   Embed   Report

Comments

Content

10186-10-89P

AID: 3487 | 27/11/2013

Draw the figure as below

First calculate all the displacements x and y as below using right angled triangle theorems
x  l cos 
y  l sin 

Where l is the length of the bar
Next calculate the potential energy of the spring
1
P.Espring  ky 2
2
Substitute the values in the above equation
1
2
P.Espring  k  l sin  
2

Next find the potential energy of the force P
P.E  2 Px
Substitute the value in the above equation
P.E  2 Pl cos 

Hence total potential energy of the system is given below
1
2
V  k  l sin    2 Pl cos 
2
Hence differentiate the above equation
dV
 kl 2 sin  cos   2 Pl sin 
d
kl 2

sin 2  2 Pl sin 
2

d 2V
0
  0 d 2
For stability at
,
. Hence differentiate the above equation and substitute
 0

d 2V
 kl 2 cos 2  2 Pl cos 
2
d

 0
Substitute
in the above equation
2
dV
 kl 2  2 Pl  0
2
d
Hence

P

kl
2

Hence the range of values of P is shown below
kl
P
2

Sponsor Documents

Or use your account on DocShare.tips

Hide

Forgot your password?

Or register your new account on DocShare.tips

Hide

Lost your password? Please enter your email address. You will receive a link to create a new password.

Back to log-in

Close