Aakash Pandita Posted September 3, 2011 Share Posted September 3, 2011 There is something that I not able to understand about force. I would like to tell it to you with an example. There is a train consisting of 1 engine of 8000kg and 5 wagons of 2000 kg each The acceleration of the train is 2 m/s2 How would you tell what force is exerted by wagon 1 on wagon 2 and why? Link to comment Share on other sites More sharing options...
swansont Posted September 3, 2011 Share Posted September 3, 2011 F=ma The net force on an object is the mass times its acceleration. The net force is made up of the vector sums of all the individual forces. Each car will have a net force on it, because they are all accelerating. It's probably best to draw a diagram and identify and label the individual forces to investigate them. Each car will feel a force from whatever is pulling on them, from both the front and back (as applicable) Link to comment Share on other sites More sharing options...
Aakash Pandita Posted September 3, 2011 Author Share Posted September 3, 2011 I am just not able to understand that when we want to find the force exerted by wagon1 on wagon2..........what mass should we multiply with the acceleration and why? Link to comment Share on other sites More sharing options...
swansont Posted September 3, 2011 Share Posted September 3, 2011 I am just not able to understand that when we want to find the force exerted by wagon1 on wagon2..........what mass should we multiply with the acceleration and why? The net force is the mass of the wagon* the acceleration; in this case 2000kg * 2 m/s^2 = 4000N Now, what are the individual forces on wagon 2? There's one from wagon 1, but are there any others? Link to comment Share on other sites More sharing options...
Aakash Pandita Posted September 3, 2011 Author Share Posted September 3, 2011 wagon 1 is joined to the engine .....and we are asked the force which it exerts on wagon 2 which is connected to wagon 3,4,5. will the force exerted by the wagon 1= mass of engine+mass of wagon 1 x a? Link to comment Share on other sites More sharing options...
swansont Posted September 3, 2011 Share Posted September 3, 2011 So wagon 1 has to pull wagons 2, 3, 4 and 5, and that force has to be present where it connects with wagon 2. What force is required to do that? Link to comment Share on other sites More sharing options...
Aakash Pandita Posted September 4, 2011 Author Share Posted September 4, 2011 f=ma so what mass should we take to multiply with acceleration...................mass of engine+mass of wagon1..........or.............mass of wagon2+3+4+5 Link to comment Share on other sites More sharing options...
swansont Posted September 4, 2011 Share Posted September 4, 2011 f=ma so what mass should we take to multiply with acceleration...................mass of engine+mass of wagon1..........or.............mass of wagon2+3+4+5 What is wagon 1 pulling? Link to comment Share on other sites More sharing options...
Aakash Pandita Posted September 4, 2011 Author Share Posted September 4, 2011 it is pulling wagons2,3,4,5 Link to comment Share on other sites More sharing options...
Sriman Dutta Posted August 12, 2017 Share Posted August 12, 2017 Suppose it's like this: F <--- |eng|= |wg1|=|wg2|=|wg3|=|wg4|=|wg5| So this is the train (or wagons). Assuming that each wagon weighs m and the engine is of mass M, then net acceleration a = F/(M+5m) So force exerted by wg2 on wg3 is = 3ma Link to comment Share on other sites More sharing options...
Country Boy Posted August 16, 2017 Share Posted August 16, 2017 The entire train has mass 8000+ 5(2000)= 18000 kg. Since it is accelerating at 2 m/s^2, the force on the entire train is 2(18000)= 36000 N. The force necessary to accelerate the engine alone would be 2(8000)= 16000 N so the cars must be exerting a retarding force on the engine of 36000- 16000= 20000 N. That, of course, means that the engine is pulling the cars with a force of 20000N. The force necessary to pull only the first car at that speed is 2(2000)= 4000 N so the first car must be pulling the second car with force 20000- 4000= 16000 N. Similarly, that second car must be pulling the third car with force 16000- 4000= 12000 N, the third car must be pulling the fourth car with force 12000- 8000= 4000 N and the fourth car must be pulling the fifth car with force 8000- 4000= 4000 N. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now