Help with mechanics question please (1 Viewer)

J-Wang · Jun 9, 2012

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 · Jun 9, 2012

does anyone know how to do this?

J-Wang · Jun 9, 2012

help please

J-Wang · Jun 9, 2012

anyone?

IamBread · Jun 9, 2012

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

J-Wang · Jun 9, 2012

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

thanks mate

IamBread · Jun 9, 2012

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 · Jun 9, 2012

IamBread said:
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

IamBread · Jun 9, 2012

J-Wang said:
the answer is 656N

I know what I did wrong....

fixed it.

J-Wang · Jun 9, 2012

IamBread said:
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 · Jun 9, 2012

J-Wang said:
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.

IamBread · Jun 9, 2012

There we go

qwerty44 · Jun 10, 2012

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

IamBread · Jun 10, 2012

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

Help with mechanics question please (1 Viewer)

J-Wang

Member

J-Wang

Member

J-Wang

Member

J-Wang

Member

IamBread

Member

J-Wang

Member

IamBread

Member

J-Wang

Member

IamBread

Member

J-Wang

Member

IamBread

Member

IamBread

Member

qwerty44

Member

IamBread

Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)