Ques: An 8000 kg engine pulls the train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 newton and the track offers a frictional force of 5000 newtons then calculate the force of wagon 1 on wagon 2.

 

Ans ( I tried) : Total weight =18,000 kg

Net force exerted= 40000 - 5000 = 35000

Accelration: 35000 / 18000 = 35 / 18 = 3.5 m/s^2

Quote

Ans ( I tried) : Total weight =18,000 kg

kg are a measure of mass not weight.

 

12 minutes ago, student_lfcs said:

Net force exerted= 40000 - 5000 = 35000

Is the net force the same between all parts of the train?

 

Ok. Please see if I am right this time:

Total mass =18,000 kg

Net force exerted= 40000 - 5000 = 35000

Accelration: 35000 / 18000 = 35 / 18 = 1.95 m/s^2

force exerted by wagon 1 on wagon 2: Accelration x Mass

Mass in this case = 4 x 2000 = 8000 kg; Accelration is same i.e. 1.95 m/s^2

So F = 8000 x 1.95 = 15555 N

1 hour ago, student_lfcs said:

Ok. Please see if I am right this time:

Total mass =18,000 kg

Net force exerted= 40000 - 5000 = 35000

Accelration: 35000 / 18000 = 35 / 18 = 1.95 m/s^2

force exerted by wagon 1 on wagon 2: Accelration x Mass

Mass in this case = 4 x 2000 = 8000 kg; Accelration is same i.e. 1.95 m/s^2

So F = 8000 x 1.95 = 15555 N

OK that looks a whole lot better.

Hopefully you feel good too since I didn't do it for you, just offered a hint.

 

One small niggle.

My calculator says that 35/18 = 1.94 not 1.95.

In fact there is no need to work this out at all.

Just leave the common acceleration* as

[math]\left( {\frac{{35}}{{18}}} \right)[/math]

Then you can multiply it all out at the end.

Very often something will cancel out and make it even easier.

[math]F = ma = 8000*\left( {\frac{{35}}{{18}}} \right)[/math]

 

 * Notice also how to spell acceleration.

Edited August 13, 20196 yr by studiot