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I have a homework question in which a car of mass $M\ kg$ is parked on a hill inclined at $25^o$ The car is facing up the hill and I am told that the wheels are $3\ m$ apart and the centre of mass is halfway between the wheels and $0.5\ m$ from ground level.

I must find the torque due to gravity on the centre of mass of the car and the torque due to the normal forces $N$ and $K$ with respect to the same origin as in the first part of the question. Then I must finally calculate what the normal forces actually are.

The way I was going to approach this was to first calculate the Weight $$W = Mg$$

Then calculate the distance from the axis of rotation to the origin,( the centre of mass). This would be the vector $\vec{r}$ $$\sqrt{(1.5)^2 +(0.5)^2}$$

Then I would know that the vector $\vec {r}$ makes an angle of $65^o$ with respect tor the line of action of the force. That would mean the lever arm would be equal to $$\vec{r} Sin(65)$$.

Then the torque would be $W\vec{r}Sin(65)$

Is this correct or am I missing something?

For the second part of the question I would just have to use the same basic method as above except change the angle and therefore change the lever arm size.

And for the last part equate the value for torque obtained in the first part to the expression obtained in the second part and solve for $N$ and $K$?

Could someone just confirm I am going the correct way with this or suggest a different approach if what I'm doing is stupid?

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  • $\begingroup$ The first thing you should do is a free body diagram. Then the solution should be right there in front of you. The FBD would lead you to balance of forces in two directions and well as the balance of moments. $\endgroup$
    – John Alexiou
    Commented Dec 2, 2014 at 13:43
  • $\begingroup$ I did try to use one but this topic is perhaps my weakest, could you maybe point out where I am wrong at the moment? $\endgroup$
    – user
    Commented Dec 2, 2014 at 13:52
  • $\begingroup$ Note that since gravity (and hence weight) goes through the center of mass there is no net torque about that point. And since the car is not moving the net torque due to the reaction forces has to be zero. $\endgroup$
    – John Alexiou
    Commented Dec 2, 2014 at 13:58

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Here is the FBD as I understand the question

fbd

The balance of forces along the incline give us

$$ C - W \sin \theta = 0 $$

The balance of forces along the normal give us

$$ A + B - W \cos \theta = 0 $$

And the balance of torques about the bottom contact point are

$$ \ell B + h W \sin\theta - \frac{\ell}{2} W \cos\theta = 0 $$

All this works out if you resolve the weight along he incline as $W \sin\theta$ and normal to incline as $W \cos \theta$.

Now use the above information to answer the questions asked. There are 3 unknowns (A, B, C)and 3 equations.

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  • $\begingroup$ That diagram cleared everything up, Now I can fully understand the situation, may I ask which program you used to generate it? $\endgroup$
    – user
    Commented Dec 2, 2014 at 14:07
  • $\begingroup$ I used powerpoint. $\endgroup$
    – John Alexiou
    Commented Dec 2, 2014 at 14:27

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