Help with mechanics question please

**J-Wang** · 9 Jun 2012, 6:25 PM

I'm sure there's a simple way of doing this, but i just cant see what it is. could someone please do this question with steps that i can follow?
thanks

The question is:

The combined air and road resistance of a car in motion is proportional to v^2, where v is its speed. When the entire engine is disengaged the car moves down an incline making an angle of inverse sin 1/30 [sin^-1(1/30)] with the horizontal, with a velocity pf 30m/s. Find the force required to drive the car up the incline with a steady speed of 24m/s, given that the mass of the car is 1200kg and g=10m/s^2

Thanks, and good luck

**J-Wang** · 9 Jun 2012, 6:57 PM

does anyone know how to do this?

**J-Wang** · 9 Jun 2012, 7:29 PM

help please

**J-Wang** · 9 Jun 2012, 9:12 PM

anyone?

**IamBread** · 9 Jun 2012, 9:19 PM

I think I got it, I'll post a solution soon.

**J-Wang** · 9 Jun 2012, 9:30 PM

Originally Posted by IamBread

I think I got it, I'll post a solution soon.

thanks mate

**IamBread** · 9 Jun 2012, 9:38 PM

When the car is going down the hill, the forces acting upon it are
$F=mg(\frac{1}{30}) - kv_d^2 \\ $The car is travelling at a constant velocity, and so $ F=0 \\ mg(\frac{1}{30}) = kv_d^2 \\ k = \frac{mg(\frac{1}{30})}{v_d^2}$

Now that we know k, we have everything we need to find the force.

The force acting on the car when it is moving up the hill is
$F=-(mg(\frac{1}{30}) + kv_u^2) \\ = -(mg(\frac{1}{30}) + (\frac{mg(\frac{1}{30})}{v_d^2})v_u^2)$

This is the force acting on it, so the force needed to overcome it is
$F = mg(\frac{1}{30}) + (\frac{mg(\frac{1}{30})v_u^2}{v_d^2}) \\ = mg(\frac{1}{30})(1+\frac{v_u^2}{v_d^2}) \\ $Subbing in our numbers, we get $ \\ F = 656$

**J-Wang** · 9 Jun 2012, 9:42 PM

Originally Posted by IamBread

When the car is going down the hill, the forces acting upon it are
$F=mgsin^{-1}(\frac{1}{30}) - kv_d^2 \\ $The car is travelling at a constant velocity, and so $ F=0 \\ mgsin^{-1}(\frac{1}{30}) = kv_d^2 \\ k = \frac{mgsin^{-1}(\frac{1}{30})}{v_d^2}$

Now that we know k, we have everything we need to find the force.

The force acting on the car when it is moving up the hill is
$F=-(mgsin^{-1}(\frac{1}{30}) + kv_u^2) \\ = -(mgsin^{-1}(\frac{1}{30}) + (\frac{mgsin^{-1}(\frac{1}{30})}{v_d^2})v_u^2)$

This is the force acting on it, so the force needed to overcome it is
$F = mgsin^{-1}(\frac{1}{30}) + (\frac{mgsin^{-1}(\frac{1}{30})v_u^2}{v_d^2}) \\ = mgsin^{-1}(\frac{1}{30})(1+\frac{v_u^2}{v_d^2}) \\ $Subbing in our numbers, we get $ \\ F = 37593N$

the answer is 656N :P

**IamBread** · 9 Jun 2012, 9:48 PM

Originally Posted by J-Wang

the answer is 656N :P

I know what I did wrong....

fixed it.

**J-Wang** · 9 Jun 2012, 10:03 PM

Originally Posted by IamBread

I know what I did wrong....

fixed it.

thanks mate. would it be possible to upload a diagram of how u did all that? just so it can help visualise. i understand the maths, just visually the diagrams help lol

**IamBread** · 9 Jun 2012, 10:40 PM

Originally Posted by J-Wang

thanks mate. would it be possible to upload a diagram of how u did all that? just so it can help visualise. i understand the maths, just visually the diagrams help lol

I didn't use a diagram lol, that's where the error came in