PRACTICE PROBLEM $2\mathrm{.}2$

SOLUTION

VEHICLE DESIGN, ENGINE$-$GENERATED TRACTIVE EFFORT, ROLLING RESISTANCE, ACCELERATION

A rear$-$wheel$-$drive car has an engine running at $3296$ revolutions per minute. It is known that at this engine speed the engine produces $80$ horsepower. The car has an overall gear reduction ratio of $10,$ a wheel radius of $16$ inches, and a $95\%$ drivetrain mechanical efficiency. The weight of the car is $2600lb,$ the wheelbase is $95$ inches, and the center of gravity is $22$ inches above the roadway surface. What is the closest distance the center of gravity can be behind the front axle to have the vehicle achieve its maximum acceleration from rest on good, wet pavement?

Note: Open boxes in equations $"$ are to be completed by the reader

This problem illustrates the relationship between engine$-$generated tractive effort $F_c,$ and the maximum tractive effort $F_{m\text{m}\text{a}\text{x}}.$ To achieve maximum acceleration, we can see from Eq$.)=({}_{m}ma$ that we want the available tractive effort $F($which is the lesser of the engine$-$generated tractive effort $F_c,$ and the maximum tractive effort $($:$F_{\text{m}\text{a}\text{x}}\}$ to be as high as possible. The problem provides information that will allow us to calculate the engine$-$generated tractive effort $F_c$ at the conditions specified. So the point of interest in this problem is to determine an $F_{\text{m}\text{a}\text{x}}$ that will enable the vehicle to use all of the available engine$-$generated tractive effort under the conditions given. It is known from Eq$.2\mathrm{.}14$ that $F_{\text{m}\text{a}\text{x}}$ is a function of the location of the center of gravity from the front axle. In this problem, if the center of gravity is too close to the front axle, $F_{\text{m}\text{a}\text{x}}$ will be less than $F_e$ and the vehicle will not be able to achieve it maximum acceleration based on it engine$-$generated tractive effort. This condition is illustrated in the lowest speed region of Fig. $2\mathrm{.}5$ where $F_e$ curve exceeds the $F_{\text{m}\text{a}\text{x}}$ line. If $F_{\text{m}\text{a}\text{x}}F{F}_{\text{m}\text{a}\text{x}}{F}_c{F}_{\text{m}\text{a}\text{x}}{F}_{\text{m}\text{a}\text{x}}=F_e{F}_{\text{m}\text{a}\text{x}}=F_e=\frac{M_e{}_0{}_d}{r}{M}_c{M}_c{F}_{\text{m}\text{a}\text{x}}=F_e=\frac{M_e{}_0{}_d}{r}=\frac{}{}=908\mathrm{.}27lb{F}_{\text{m}\text{a}\text{x}}{l}_f=0\mathrm{.}9F_{\text{m}\text{a}\text{x}}=\frac{W\frac{l_f-f_{r}h}{L}}{1-\frac{h}{L}}{f}_{tb}V=,f_{rt}=0\mathrm{.}01(1+\frac{V}{147})=0\mathrm{.}01(1+\frac{}{147})=l_f=29\mathrm{.}42$