# Homework Help: Tractive effort of car D'Alemberts principle

1. May 27, 2010

Hi

I have been working on this problem now for a good few weeks and I think I am nearly there just need help on the last bit.

The Question:

A car of mass 1200kg accelertaes from 2.5 to 5.0 ms while travelling up a slope of 1in 10, through a distance of 60m. If resistance to motion is 105N determine using both an energy method and D'Alemberts principle

a) The tractive effort F required
b) Work done
c) average power required to accelerate load

What I have done so far.

Part a)

I worked out acceleration using a=(v^2 - u^2 ) / 2s giving me a= 0.15625 m/s

Then using m x a to work out inertia resistance = 187.5N

I then worked out the angle ( bottom left of traingle ) by using slope of 1 in 10 = 0.1 and doing tan-1(0.1 to give 5.71 degress

I then worked out the gravitational force m x g x sin pheta ( 1200 x 9.81 x sin(5.71 )

This gave 1171.357N

Adding these with my original resistance to motion gave me a tractive effort of 1463.857139N

Part a) done hopefully

Part b)

Work done = f x s
1463.857139 x 60 = 87831.42834 Nm

Part c) Power = work done / time taken

Time taken = change in velocity / acceleration

2.5/0.15625 = 16 secs

Power = 87831.42834 / 16 = 5489.4642

I divided this my distance of 60m to give 91.49107 Nms

This is where I am stuck I understand I need to use conservations of energy which mean working out PE and KE.

Intial KE

KE= 1/2 m x v^2

600 x 2.5^2 = 3750J

Final KE

600 x 5^2 = 15000

Give me a change in KE of 11250

And that as much as I can do.

Sorry for it being quite long but I really want to get through this question. Any input is greatly appreciated. Thanks

I will attached a picture of a diagram that also comes with the questions

Thanks again

#### Attached Files:

• ###### Untitled.jpg
File size:
12 KB
Views:
593
2. May 27, 2010

### PhanthomJay

badsanta010, welcome to PF! You've done some good work thusfar using Newton's laws, but with a clarification required and an error, see below. (Note: You should probably round off your answers to 2 significant figures)
This looks good so far, very good
Nm is usually referred to as Joules, but this is where clarification is needed. This is the work sone by the tractive force, which I assume the problem is asking for? Most all the forces (tractive force, gravity, resistive forces) do work, so the problem should be more specific when it asks for "work" done.
yes, this is the average power delivered by the car, units should be Joules/s, or Watts
why? You already calculated the average power above. Go no further.
Using energy methods, your change in KE is correct. What's the change in PE? Then use the conservation of energy equation which states that the change in KE plus the change in PE must equal the work done by the non-conservative (tractive and resistive) forces.

3. May 28, 2010

Hi

Thanks for the help

I cannot clarify on the work done. I assume as you said it must be asking for work done of the tractive effort

I knew work done = change in KE

I did'nt know change in PE could also be added to this

Well thanks again, I will try and give this a go when I am free over the weekend and hopefully I may of cracked it.

Thanks

4. May 28, 2010

### PhanthomJay

Total (or net) work done by ALL forces is change in KE, that's why I asked for clarification...I woudn't use this equation
For some clarification (or possible confusion), note that W_net = change in KE, and W_nc = change in KE plus change in PE , where W_net is the work done by all forces, including gravity, and W_nc is the work done by non-conservative forces (that excludes gravity).