Attatchment depicts a truck which has a centre of mass C

[framehead]

Hello. Could someone possibly help me with this question?
1. The problem:

The attatchment depicts a truck which has a centre of mass C.

• The truck is front wheel drive.
• The geometrical position of C is defined.
• The coefficient of friction between the tyres and the road is μ.
• Neglect the moment of inertia of the wheels.

The driver wants to find the maximum acceleration of the vehicle in the positive x direction.
Show the maximum acceleration of the truck is given by;

http://www.imagefilez.com/out.php/i279772_13.jpg

I have no idea how to do this question but here is my attempt at a solution:
\SigmaM_b=0; mgb-N_a(a-b)=0
\SigmaF_y=0; N_b+N_a-Mg=0
\SigmaF_x=0; ah-\muh=0

and that's about where I'm at at the moment and I know I'm totally wrong! I would appreciate anybodys help.

 

[moemoney]

For there to be a maximum acceleration, doesn't there have to be a maximum \SigmaFx value? Meaning, shouldn't \SigmaFx not be 0?

 

[PhanthomJay]

Your equations need some work for sure. The Fy forces add to zero, but the Fx force is not zero, as moemoney notes. When the truck accelerates, the force that causes the acceleration is the static friction force between the front (front wheel drive) tires and the road. Without this friction, the truck could not move. Because the c.m. of the vehicle is above the tires, this force at road surface tends to rotate the vehicle about the c.m. in a counterclockwise direction, causing the vehicle to lift off the front tires and resulting in less normal force ( and less available friction force) on those tires. To solve this problem, a bit of advanced knowledge is necessary. You should replace the friction force at the front tires with an equivalent force and ccw couple acting at the c.m. Then calculate the normal force at the front tires due to both the vehicle weight and the couple, and apply Newton 2 at the c.m. to solve for x double dot.

 

[nvn]

framehead: d'Alembert's principle says, the mass times acceleration can be represented as a horizontal force F_cx acting to the left at point C. There is also a horizontal, frictional force that the road applies to the front tyres, F_bx, acting to the right. And, as you know, there is a vertical force m*g at point C, and a vertical force N_b at point b. Now perform summation(F_x) = 0, and summation(M_a) = 0.

As the acceleration increases, N_b decreases. The maximum possible acceleration is when finally F_bx = mu*N_b.

 

[framehead]

I appreciate everybodys help but i still seem to be getting nowhere. This is what I've got from all the advice given:
\SigmaM(x)_c=0=R₁a-R₂b+\muh
and
R₂(reaction at front wheel)=mga+\muh/a-b

 

[PhanthomJay]

I appreciate everybodys help but i still seem to be getting nowhere. This is what I've got from all the advice given:
\SigmaM(x)_c=0=R₁a-R₂b+\muh

your last term \mu h is incorrect. The moment about the cm of the friction force at R2 is the friction force times h. What is the friction force in terns of R2? Also, watch your signs. If clockwise is positive, stick to that convention.

and
R₂(reaction at front wheel)=mga+\muh/a-b

I don't know where this comes from. You should sum forces in the y direction to get your 2nd equation. With these 2 equations, you can solve for R2, and then get the friction force, and then solve for the acceleration in the x direction.