Mechanics Probem (1 Viewer)

iLearn4U · Jun 29, 2015

Hi, I am new here and need help with this problem.

The combined air and road resistance of a car in motion is proportional to v^2, where v is its speed.

When the engine is disengaged the car moves down an incline making an angle of sin^-1 (1/30) with the horizontal, with a velocity of 30 m/s.

Find the force required to drive the car up the incline with a steady speed of 24 m/s, given that the mass of the car is 1200 kg.

Thank you

Drsoccerball · Jun 29, 2015

iLearn4U said:
Hi, I am new here and need help with this problem.

The combined air and road resistance of a car in motion is proportional to v^2, where v is its speed.

When the engine is disengaged the car moves down an incline making an angle of sin^-1 (1/30) with the horizontal, with a velocity of 30 m/s.

Find the force required to drive the car up the incline with a steady speed of 24 m/s, given that the mass of the car is 1200 kg.

Thank you

I got to :
$F= 19.2( \frac{1200}{r}-1) -392)$
Assuming g= 9.8

InteGrand · Jun 29, 2015

Drsoccerball said:
I got to :
$F= 19.2( \frac{1200}{r}-1) -392)$
Assuming g= 9.8

What's r? This isn't a circular motion (banked turn) Q.

Drsoccerball · Jun 29, 2015

InteGrand said:
What's r? This isn't a circular motion (banked turn) Q.

Isnt it ahahah ?

InteGrand · Jun 29, 2015

Drsoccerball said:
Isnt it ahahah ?

It's essentially just an inclined plane problem.

iLearn4U · Jun 29, 2015

Yes, its a banked track problem. I don't know what to do with the 30 m/s and 24 m/s.

The answer is 656 N.

leehuan · Jun 29, 2015

Couldn't be bothered doing the questions so here's the diagrams. I reckon there's resisted motion and 'banked tracks' here. There are vectors but we don't have a centripetal force component as the car is travelling up, not into the paper.

Edit: That should not say "thurst"

Also, something tells me that the 'initial conditions' enable us to find k..

InteGrand · Jun 29, 2015

$This isn't banked turn I'm pretty sure. This is just inclined plane (i.e. moving up and down the plane). Banked turns are essentially when you take a turn in a circle that is banked.$

$To find $k$ (note that its units will be $\text{N}\text{ m}^{-2}\text{ s}^2$) note that initially, when the velocity is a constant $v_1=30\text{ m s}^{-1}$ down the plane, since the sum of forces must be zero (by Newton's Second Law, since the velocity is constant), we have $mg\sin \theta = kv_1^2$, since $mg\sin \theta$ is the magnitude of the force acting down the slope, and $kv_1^2$ is the magnitude of the frictional forces acting up the slope (also we are given that $m=1200\text{ kg},\sin \theta = \frac{1}{30}$, and we use $g=9.8\text{ m s}^{-2}$).$

$\therefore k=\frac{mg\sin\theta}{v_1^2}$

$=\frac{1200\text{ kg}\times 9.8\text{ m s}^{-2} \times \frac{1}{30}}{(30\text{ m s}^{-1})^2}$

$$=\frac{98}{225}\text{ N}\text{ m}^{-2}\text{ s}^2$.$

$Now, when the velocity is a constant $v_2=24\text{ m s}^{-1}$ up the plane (so the frictional forces point $\emph{down}$ the plane, since frictional forces always oppose the motion), the force up the plane is $F_\text{car}$ (which we want to find), and the forces down the plane are the component of gravity $mg\sin \theta$ and frictional forces $kv_2^2$. Since these forces must balance again, we have$

$F_\text{car}=mg\sin \theta + kv_2^2$

$=1200\text{ kg}\times 9.8\text{ m s}^{-2} \times \frac{1}{30} + \frac{98}{225}\text{ N}\text{ m}^{-2}\text{ s}^2\cdot (24\text{ m s}^{-1})^2$

$=(1200\times 9.8 \times \frac{1}{30}+\frac{98}{225}\times 24^2)\text{ N}$, as $\text{ kg m s}^{-2} = \text{ N}$

$$=642.88\text{ N}$.$

$If we used $g=10\text{ m s}^{-2}$, we would get $k=\frac{4}{9} \text{ N}\text{ m}^{-2}\text{ s}^2$ and $F_\text{car} = 656 \text{ N}$.$

iLearn4U · Jun 30, 2015

The answer in the back of the book is 656 N.

We have to use g=10 m/s^2 and F will equal to 656 N.

Thanks for your response guys/girls. I appreciate it.

Mechanics Probem (1 Viewer)

iLearn4U

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Drsoccerball

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InteGrand

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Drsoccerball

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InteGrand

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iLearn4U

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leehuan

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InteGrand

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iLearn4U

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