# What position of the center of gravity can make the front wheels of the car lift off the ground?

I have a question regarding the position of center of gravity required to just lift off the front wheel of a vehicle

Consider a vehicle of mass $m$ having a center of gravity at height $h = 0.5m$ from the ground. The coefficient of friction between the tire and the ground is 1. Assume that the engine supplies just enough torque to utilize all the friction force without causing the wheels to spin.

My question is where should the center of gravity of the vehicle should be located in relation to the rear wheels to make the wheels just lift off the ground. I have solved the question as shown, and I get a 'NEGATIVE' value for L2 meaning that COG should be 0.5m behind the rear wheel, but the solution says that it should be 0.5 m infront of the rear wheels.

Can someone help me out on this! Since, my handwriting is not clear, I am writing the equations here too

Equilibrium in vertical direction

(1) $N_1 + N_2 = mg$

Equation of motion in horizontal direction

(2) $F_{tr} = ma$

Here $F_{tr}$ is the traction force on the rear tire which propels the vehicle. Also,

(3) $F_{tr} =\mu N_2$

The balance of the torque on the rear tire $\to$ the net moment on the rear tire about the contact point is zero

(4) $mgl_2 + 0.5 \ ma = 0$.

Now, since the vehicle must just lift off the ground

(5) $N_1 = 0; \ N_2 = mg$.

Using the equations (2), (3), and (5),

(6) $a = \mu g$.

Now using the equation (6) in the equation (4),

(7) $l_2 = -0.5 \ \mu = -0.5$

Now, the final value of $l_2$ is negative, which means that it is opposite to the assumed direction. So the center of mass should be 0.5 m behind the rear wheels.

The only difference between my method and the solution manual which I am referring to is that they have considered inertia force(pseudo force) on the vehicle, and thus, they get the answers $l_2 = 0.5$, which means 0.5 m in-front of the rear wheels

• Note: Use $...$ for variables to show with ${math}$ notation instead of '...'. See physics.stackexchange.com/help/notation Jan 12, 2015 at 19:08
• Can't you edit the calculus? It's difficult to read your handwrite. Jan 12, 2015 at 19:17
• @Sofia - I have updated my question and included the equations. Jan 12, 2015 at 21:11
• @ja72 - It's modified now. Thanks for the suggestions Jan 12, 2015 at 21:11
• Note $$N_1+N_2=mg$$ to get centered equations and $$N_1+N_2=mg\tag{1}$$ to get a centered equation with a right-adjusted label. This will make it a bit more readable. Jan 12, 2015 at 21:15

We often use the sum of torques about an axis to find the case of zero rotation by finding where the sum is zero. The reason this works is that the case of no rotation must also be a case of no rotational acceleration and therefore no change in angular momentum.

$$\tau_1 + \tau_2 + .... = I \frac{\Delta L}{t} = 0$$

But in your case the car is accelerating. The angular momentum of your system is $$L = d \times v$$ The offset distance between the center of mass of the car and the axis is fixed ($0.5m$), but the velocity of the car is changing. That means the angular momentum is also changing and there is a net torque. You cannot solve for the case where net torque is zero.

In the frame of the car, the axis is no longer accelerating, but you have a fictitious force on the center of mass to account for.

• I have taken torque about center of mass to be zero since there is no pitching moment in the vehicle. I have posted an update answer to the problem:) Jan 13, 2015 at 0:43
• Right. When you move the axis to the center of mass, the angular momentum of the center of mass becomes zero and you can ignore the acceleration. Jan 13, 2015 at 0:46

I think I understood the reason. I am incorrect in taking the torque of ma directly. This is because ma is the result of the forces acting on the system. Instead I should have balanced the torque about the center of mass, with the equation [4] looking like this $$mg∗l2−Ftr∗0.5=0$$ [Torque balance about center of mass, assuming that there is no pitching moment]

Now I can replace $$Ftr=m∗a$$

which on further solving will give

$$l2=.5$$

It's just a simple sign-error.

You wrote

$$mgl_2+~ma0.5=0~.$$

But when you accelerate the car to the right then you get a reaction force $ma$ which points in the other direction, so your $m \vec a$-arrow should point to the left. The equation then reads

$$mgl_2-~ma0.5=0~.$$

Then you have your torque balance about the rear wheel and $l_2=0.5$ at the end.

• But if you look at the situation from a Newtonian (non-accelerated) frame of reference $a$ will be always pointing to the right (which I have assumed as positive). Jan 12, 2015 at 21:22
• Yes you're right. I answered too quick.
– user42076
Jan 12, 2015 at 21:28
• That's okay. Even if you think by intuition you will get the fact that if the acceleration is producing a clockwise moment, then the weight should produce an anticlockwise moment to balance it out. For that the center of mass should be behind the rear tire. I am bit puzzled over the question! Jan 12, 2015 at 21:31
• @MasoomKumar Now I got it! Okay we're in a newtonian frame. You accelerate the car. So you have an acceleration to the right at the rear tire, so a overall force to the right at the rear tire. At the same time, since action=reaction you have a (non fictitious) reaction force with the same magnitude at the center of mass. So you have one ma-vector to the right at the rear tire and one ma-vector to the left at the center of mass. Since the rear tire is the pivot point the ma-vector to the right doesn't produce a torque, only the weight and the ma-vector to the left do. Do you agree?
– user42076
Jan 12, 2015 at 21:41
• Thanks for the explanation. But I would gently disagree with you at this point. According to my understanding irrespective of wherever the force acts on the body, when looking from a Newtonian frame of reference $ma$ will act only at the center of mass. Only if we sit inside the accelerating vehicle, do we feel that we are pushed back, and that necessitates the use of a fictional force. Jan 12, 2015 at 21:51

Simple answer to your title question: the forward wheels will lift off the ground when the total vehicle center of mass lies behind the hinder axle (i.e. the vertical projection of the COM onto the ground lies behind the vertical projection of the hinder axle onto the ground). Otherwise there must be a nett reaction force on the forward wheels to counter the torque about the hinder axle of the weight; the latter always acts through the center of mass.