Motion law for constraints
Constraints motion best example is in your childhood your father hold your hand and walk on the road means your motion was guided by your father as your father will move you will follow the path of your father hence your motion is called constraint motion.
if two block A & B are attached together and we are applying force on block A then block A & B will move together then we can say displacement of block A & B will be same because both are attached moving together from here we can write xa = xb where xa displacement of block A and xb displacement of block B now differentiate this equation with respect to time we will get
d(xa)/dt = d(xb)/dt
va = vb ( velocity of block A & B are also same ) again differentiate with respect to time
d(va)/dt = d(vb)/dt ( acceleration of block A & B are also same)
hence these equation are called constraints equation.
above are simple pulley question where we solve easily but for complex question we need some important method to solve question so one method is virtual work method by which we use to solve pulley constraints question see below picture.
Now we will learn virtual work method for solving difficult problem virtual work method made easy to understand complex problem you can use this method for complex problem suppose two block are attached through single pulley as shown in figure.
From above second picture lets big pulley has tension T in rope big pulley is fixed as shown now small pulley has tension T1 in rope small pulley is movable its mean that small pulley will work but big pulley is fixed hence it will not work now apply work done by each pulley so for big pulley suppose small pulley moves up by displacement xp now we can write equation for big pulley rope
T*xa+T*xp = 0 ( Net work done by rope is zero )
xa +xp = 0
now for small pulley since pulley is massless hence tension in rope will be T/2 small pulley upward tension in rope is T hence downward tension in rope will be T/2
now apply work done
-T/2*xb -T/2*xp -T/2*xp = 0
-xb -xp-xp = 0
xb = -2xp or xp = -xb/2 put in above equation we get.
xa -xb/2 = 0
xa = xb/2 ( this is our constraint equation relation )
now we can write this equation in short form also work done by rope first and work done by rope second will be zero w1+w2 =0
(T*xa+T*xp) + (-T/2*xb-T/2*xp-T/2*xp) = 0
see here pulley term is going to cancelled its mean that if we don’t write pulley work or forget then result will be correct this is important point to remember
xa = xb/2 i hope that you have enjoyed learning Motion law for constraints and your concept is now clear thanks for reading.
dated 12th Aug 2018