Analysis Poses Difficulty

Homework? It sounds like it.

Are these covered wagons or buck board type

Unfortunately, there is no way to circle just two wagons. So, you will be captured by Indians (American, not east Indians) before the experiment can be completed.

They will torture you by making you do long division in your head while you are tied to a wagon wheel with a light, inextensible chain.

Pls don't embarass me further. That's why I even posted this anonymously.pls I need ur assistance. A little sacrifice.

Convert everything in simple SI units.

calculate the nett force on Body1 (the 3Ton Body) when it is greater than zero it starts moving. (Of course it is not important to do this step... why?)

Calculate the total force on the system (it will be a function of time)

When it becomes positive the second body starts moving.

or

Do a force balance on bodies separately (Both will have the string tension as the unknown parameter)

Calculate the tension on the string.

It was simple, I calculated just now.

BTW- Don't forget to use Newton's second and third laws.

are these wagons pulled by horses or a mule team?

The way this threads going it's probably one of Dr. Dolittle's "Push me Pull you's".

1. Draw a FBD (Free Body Diagram)
2. Label all forces and other pertinent bits of data
3. apply newton's laws
4. Report back

To solve such a problem you MUST first understand what happens from the physical point of view after you got it writing the equations is easy.

Phase #1: chain is not tensioned so that the 3T wagon starts to move, you have all what you need to analyze this phase so that you can do it further on your own.

Phase #2: the chain is tensioned and the second wagon offers as well an inertial as a rolling resistance. The force with which the chain acts upon is less/= 10 kN. This force will accelerate the second wagon. But at same time the first wagon will encounter same force as added resistance so that its movement is less accelerated. As long as the second wagon has not reached the 1st wagon's speed the chain will slip which gives you a hint about the accelerating/resistance force

Phase #3: the chain will not slip any more when both wagons have same speed. After this point the 2 wagons can be considered as a single unit with the mass of both and the rolling resistance of both.

This is the way to think when you are confronted with such problems: try to understand the physics and then all is simple! Send you detailed results for check after you wrote all equations and made all computations.

Thanks guys. I have been trying this thing. The physics I need to get is not FBD but " a light inextensible chain which is rigidly attached to one wagon and passes through a friction grip on the other so that if the pull exceeds 10kN the chain will slip" as opposed to what Nickname said about the chain slipping until both wagons reach the same velocity.

Seriously, why can you not just ask your teacher/instructor/professor for help?

I couldn't resist the "tie you up and make you do long division in your head" thing.

Did you read #5 and #7?

Either you presented an erronated text or you do not read it or you do not understand it. If the chain is tensioned all time then the 2 wagons are connected in tension to each other from the start and there will never be a slip, both would be accelerated with the pulling force minus rolling resistance and the acceleration has to consider all mass (ie sum of both). The fact that you do not take into consideration an important part of the problem definition shows that you did not understand the physics. I would suggest you really go to your teacher and ask him for a couple of hours for bringing you up with problem reading and understanding. Question: why a bond will slip ? Answer: because the velocity difference is not zero since when both mobiles move at same speed there is no reason to slip, the rolling resistance is a lot lower than the threshold. THINK AGAIN may be you will get the point.

Thanks. I GET IT NOW. Thanks again.

It would be good for you to explain what you got and send the equations you wrote with obtained results.

Actually I thought I had gotten it but I haven't. The thing is this problem was taken from Mechanics of Machines (Advanced Theory and Examples) by J. Hannah and R. C. Stephens. It's not homework. Its just that I took it upon myself to develop my analytical skills in mechanical engineering. So I decided that I should solve all the problems in the textbook and I cannot go on to other problems unless I get this one solved. The question is a standard University of London question so I don't think there is any error in it. However, I will quite appreciate it if u send an attachment of ur own analysis to me at xandro101@yahoo.com this will enable me carry on with my project and I will learn a fundamental principle I may be missing out. From the textbook, the answers given are 0.428m/s for velocity after the chain has stopped slipping and 0.46s for the time of slip. Pls u can send ur solution as a picture file. Thanks in advance.

My answer is a clear "NO" you want to progress, very good I appreciate. You got from me ALL explanations to make you able to solve it. If you cant then go back to the books and work your path through till you came to an explanation which you send here and get corrections if you are wrong but FIRST you have to work, precheewed food is not good for your future. I give a last hint, a problem has "entry" elements make a list of ALL what is given as information in the problem text ad try to understand the possible reason for every "entry". Then try to explain to your self how you can use those entries and let you guided by the phases I described. Good luck