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enerj
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http://build1.net/rand/car.bmp
Given the variables in the drawing, and coefficient of friction u, what is the maximum possible acceleration if the car is rear wheel drive? Neglect rotational inertia of the wheels.
Up and right is (+) for forces along with counter-clockwise being (+) for moments
[tex]\Sigma[/tex]F = m * a
[tex]\Sigma[/tex]Mc = 0
I have established the following variables in my attempt to find the solution;
W - vehicle weight
Nf - normal force on front wheel
Nr - normal force on rear wheel
Ff - frictional force
I gave the car a weight, W, applied downward at the center of mass. From this, a normal force is applied upward at the front and rear wheel, Nf and Nr, respectively.
There is also a frictional force opposing the cars forward motion, Ff pointing to the left.
Summing the forces in the y direction yields W = Nf + Nr
And the moment equation is
[tex]\Sigma[/tex]Mc = 0 = -(Nr * A) + (Nf * (B - A)) + (Ff * Y)
I am pretty sure that the force the tire applies to the ground will have to be at the verge of breaking friction for maximum acceleration, or u * Nr
Unfortunately I could not relate the previous equations and would appreciate a pointer. By the way, the solution is
[(B - A) g * u ] / [(B - uY)
where g is gravity.
Thanks guys
Homework Statement
Given the variables in the drawing, and coefficient of friction u, what is the maximum possible acceleration if the car is rear wheel drive? Neglect rotational inertia of the wheels.
Homework Equations
Up and right is (+) for forces along with counter-clockwise being (+) for moments
[tex]\Sigma[/tex]F = m * a
[tex]\Sigma[/tex]Mc = 0
The Attempt at a Solution
I have established the following variables in my attempt to find the solution;
W - vehicle weight
Nf - normal force on front wheel
Nr - normal force on rear wheel
Ff - frictional force
I gave the car a weight, W, applied downward at the center of mass. From this, a normal force is applied upward at the front and rear wheel, Nf and Nr, respectively.
There is also a frictional force opposing the cars forward motion, Ff pointing to the left.
Summing the forces in the y direction yields W = Nf + Nr
And the moment equation is
[tex]\Sigma[/tex]Mc = 0 = -(Nr * A) + (Nf * (B - A)) + (Ff * Y)
I am pretty sure that the force the tire applies to the ground will have to be at the verge of breaking friction for maximum acceleration, or u * Nr
Unfortunately I could not relate the previous equations and would appreciate a pointer. By the way, the solution is
[(B - A) g * u ] / [(B - uY)
where g is gravity.
Thanks guys
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