Tractive effort using D'alemberts principles

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SUMMARY

The discussion focuses on calculating tractive effort using D'Alembert's principles for a vehicle with a mass of 750 kg accelerating up a 1 in 10 incline. The vehicle accelerates from 20 km/h to 60 km/h over 8.5 seconds, facing a frictional resistance of 0.5 kN. Key calculations include determining tractive effort, work done, and average power developed, while emphasizing the need for correct unit conversions and energy equations. The discussion highlights common errors in applying energy methods and the importance of understanding the distinction between net average power and engine power.

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  • D'Alembert's Principle in mechanics
  • Kinetic and potential energy equations
  • Unit conversion from km/h to m/s
  • Basic principles of work and power calculations
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  • Explore the differences between net work and engine power in mechanical systems
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Hello this is my last questions i have to finish the year, the problem I'm having is finding the correct way to work it out. I've looked at others but there their problem has differences and what units to use.

Homework Statement



A vehicle of mass 750kg accelerates up an incline of 1 in 10 (sine) increasing its speed from 20 kmh^-1 to 60 kmh^-1 in a time of 8.5 second; if the frictional resistance to motion is 0.5kN. making use of D'alembert's principles; Determine:

1. the tractive effort between the driving wheels and the road surface.
2. the work done during the period of acceleration
3. the average power developed
4. use energy method to confirm your answers in 2.

Homework Equations



1. tractive effort = change in ke + change in pe
pe= mgh
ke= 1/2MV
initial ke = mass/2 x v(intial time)^2
final ke = mass/2 x v(end time)^2
time taken = change in velocity/acceleration
work done = fs
power = work done/time taken

The Attempt at a Solution



1. tractive effort

initial Ke = 750kg/2 x (no initial time so i used 1) = 375
final Ke = 750kg/2 x 8.5^2 = 27093.75
change in Ke = 27093 - 375 = 26718

Pe = 750 x 9.81 x 40sin( I'm not sure how to get height

2. work done = f x (20kmh - 60kmh) = f x 40kmh
acceleration = v^2(final time) - (time taken)8.5^2/2s
f = 750kg x acceleration

3. power = work/8.5s

Thank you very much for any help in advanced.
 
Last edited:
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Ouch, you are making all sorts of errors. Are you familiar with D'Alembert's Principle? If not, at least use Newton's laws, although the problem asks otherwise; anyway, your only attempt was (incorrectly) using energy methods.

Tractive effort is the tires driving force that acts uphill (gravity and friction act downhill). You have shown it as an energy change, which is wrong.
Check your equation for Kinetic Energy!
Convert km/hr to m/s for consistency in units.
You have different equations for work done. Check your change in energy equation .

It is also unclear whether the problem is looking for net average power or just the average power delivered by the engine. I'm guessing net work (total work) done.

You've got to start again from scratch on this one. There are lots of concepts you are missing or not applying correctly.
 

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