Question

a 8000 kg engine pulls a train of 5 kg wagons , each of 20000 kg along a horizontal track. if the   engine  exerts a force of 40000 n and the track offers a friction of 5000 n . then calculate

(a) the net accelerating force

(b) the acceleration of train and

(c) the force of wagon 1 on wagon 2

Answer 1

a) Net Force  = Engine Force - Frictional force
                 = 40000 Newtons - 5000 Newtons = 35000 N
   this net force is accelerating force in the direction of movement of engine

b) accleration of train =  1/ (total mass of train) *  net Force acting on train
                             =    35000 / (8000 + 5 * 20000) = 35/108 = 0.324 meters/sec²

c) The friction force is expressed as Mu * Normal force exerted by tracks.  Mu is
     called coefficient of friction.     Normal force is equal to the weight of train.
     Assume g = acceleration due to gravity = 9.8 m/sec²

         total frictional force = Mu * (total weight of train)
         5000 Newtons = Mu * (108000 * 9.8 ) 
         Mu = 5/108*9.8 = 0.00472

Wagon1 is pulling Wagon2 with a force F1, which makes  Wagon2, wagon3, wagon4,   Wagon5 all  move with  an  acceleration of 0.324 m/sec², against the friction, that is Inhibiting wagon2 to wagon5.  Think of Wagon2 to Wagon5 as one object attached to Wagon1.  F1 is acting on the link between Wagon1 and wagon2.

       frictional force on Wagon2 to wagon5 = Mu * their weight 
                             = 0.00472  * 4 * 20000 * 9.8 = 3703.703 N

So force equation for Wagon2 (in horizontal direction) is
      [Mass of wagon2 to wagon5] * acceleration = force by wagon 1  - [ total frictional force on wagon2 to wagon5 ]
     4 * 20000  * 0.324 = Force by wagon1 - 3703.703
   Force by wagon1 on wagon2  =    29623.703  Newtons