displacement

3
First calculate all the displacements as below using the angles θ and   4tan  x θ =  B sin  y x  β = Substitute for x in the above equation B  4 tan sin  y  θ β = Substitute the value of    in above equation B  4 tan sin 30 2tan  y  θ θ = = o  Next find the distance between  A and B 4 cos  AB θ =  hen x=0, the distance between A and B is given below ( ) 0  4 m  AB  = !ence the extension of spring  s is given below ( ) 0 4 4 cos  s AB AB θ = =   Next calculate the potential energ" of the spring 2 spring # $ 2  P E ks = Substitute the values in the above equation 2 spring 2 # 4 $ % &N'm 4 2 cos # 2 40 # cos cos  P E θ θ θ   = × × ÷    = + ÷   Next find the potential energ" of the cart cart B $ (% &N  P E y = × Substitute the value in the above equation cart $ (% &N 2 tan #%0tan  P E  θ θ = × = !ence total potential energ" of the s"stem is given below 2 # 2 40 # #%0 tan cos cos V  θ θ θ   = + ÷   )sing the minimum potential energ" principal 0 dV d θ = !ence differentiate the above equation and equate it to *ero

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displacement

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7/17/2019 Displacement

http://slidepdf.com/reader/full/displacement-568e682b05fe1 1/1

First calculate all the displacements as below using the angles θ and  4tan x   θ =

 

B sin y x   β =

Substitute for x in the above equation

B   4 tan sin y   θ β =

Substitute the value of    in above equation

B   4 tan sin 30

2tan

 y   θ 

θ 

=

=

o

 Next find the distance between A and B

4

cos AB

θ =

 

hen x=0, the distance between A and B is given below

( )0

  4 m AB   =

!ence the extension of spring s is given below

( )0

44

cos

 s AB AB

θ 

= −

= −

  Next calculate the potential energ" of the spring

2

spring

#$

2 P E ks=

Substitute the values in the above equation2

spring

2

# 4$ % &N'm 4

2 cos

# 240 #

cos cos

 P E θ 

θ θ 

 = × × − ÷  

 = + − ÷  

 Next find the potential energ" of the cart

cart B$ (% &N P E y= ×

Substitute the value in the above equation

cart$ (% &N 2 tan

#%0tan

 P E    θ 

θ 

= ×

=

!ence total potential energ" of the s"stem is given below

2

# 240 # #%0 tan

cos cosV    θ 

θ θ 

 = + − − ÷  

 )sing the minimum potential energ" principal

0dV 

d θ =

!ence differentiate the above equation and equate it to *ero

3 2

2sin 2 sin40 #%0 tan 0

cos cos

## sin #$+(%

cos

θ θ θ 

θ θ 

θ θ 

 − − = ÷  

 − = ÷  

)se trial and error method to determine the angle

(0$4,θ   =o

Substitute the value of  in the expression for x

tan x   θ =

4 tan (0$4,

##$2( m

 x   =

=

o

!ence the value of x is as given below

##$2( m x   =