1. ## Help with mechanics question please

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

2. ## Re: Help with mechanics question please

does anyone know how to do this?

anyone?

5. ## Re: Help with mechanics question please

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

6. ## Re: Help with mechanics question please

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

7. ## Re: Help with mechanics question please

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
$F = mg(\frac{1}{30}) + (\frac{mg(\frac{1}{30})v_u^2}{v_d^2}) \\ = mg(\frac{1}{30})(1+\frac{v_u^2}{v_d^2}) \\ Subbing in our numbers, we get \\ F = 656$

8. ## Re: Help with mechanics question please

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
$F = mgsin^{-1}(\frac{1}{30}) + (\frac{mgsin^{-1}(\frac{1}{30})v_u^2}{v_d^2}) \\ = mgsin^{-1}(\frac{1}{30})(1+\frac{v_u^2}{v_d^2}) \\ Subbing in our numbers, we get \\ F = 37593N$

9. ## Re: Help with mechanics question please

Originally Posted by J-Wang
I know what I did wrong....

fixed it.

10. ## Re: Help with mechanics question please

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

11. ## Re: Help with mechanics question please

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
But I'll draw something up.

12. ## Re: Help with mechanics question please

Mechanics help.png
There we go

13. ## Re: Help with mechanics question please

Mechanics looks hard, though everyone says its not compared to others. Can't wait to start MX2.

14. ## Re: Help with mechanics question please

It's not hard, it is fun though, was probably my favourite topic in MX2! When you first look at it, it looks hard, but when you do it it's not really

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