Centers of Mass of Vehicles A car is placed on a platform that measures the normal force exerted by each tire independently (Fig. 7.37). Measurements made with the platform horizontal and with the platform tilted at α = 15° are shown in Table 7.9. Determine the position of the car's center of the

Question

Centers of Mass of Vehicles

A car is placed on a platform that measures the normal force exerted by each tire independently (Fig. 7.37). Measurements made with the platform horizontal and with the platform tilted at α = 15° are shown in Table 7.9. Determine the position of the car's center of the car's center of mass.

Table 7.9 Measurements of the normal forces exerted by the tires

Wheelbase = 2.82 m
Track = 1.55 m	Measured Loads (N)
	α = 0	α = 15
Left front wheel,[latex]N_{LF} [/latex]	5104	4463
Right front wheel,[latex]N_{RF} [/latex]	5027	4396
Left rear wheel,[latex]N_{LR} [/latex]	3613	3956
Right rear wheel,[latex]N_{RR} [/latex]	3559	3898

Accepted Answer

We draw the free-body diagram of the car when the platform is in the horizontal position in Figs, a and b. The car's weight is

[latex]W = N_{LF} + N_{RF} + N_{LR} + N_{RR} = 5104 + 5027 + 3613 + 3559 = 17.303 N[/latex]

From Fig. a, we obtain the equilibrium equation

[latex] \sum{M_{(z axis)} } + (wheelbase)(N_{LF} + N_{RF}) - \overline{x}W = 0 ,[/latex]

which we can solve for [latex] \overline{x}[/latex] :

[latex]\overline{x} = \frac{(wheelbase)( N_{LF} + N_{RF} )}{W} = \frac{(2.82)(5104 + 5027)}{17.303} = 1.651 m.[/latex]

From Fig. b,

[latex]\sum{M_{(x axis)} } = \overline{Z} W - (trak)(N_{RF} + N_{RR} ) = 0 ,[/latex]

which we can solve for [latex] \overline{z}[/latex] :

[latex] \overline{z} = \frac{(trak)(N_{RF} + N_{RR} )}{W} = \frac{(1.55)(5027 + 3559)}{17,303} = 0.769 m. [/latex]

Now that we know [latex] \overline{x}[/latex], we can determine [latex] \overline{y}[/latex] from the free-body diagram of the car when the platform is in the tilted position (Fig. c). From the equilibrium equation

[latex]\sum{M_{(z axis)} } = (wheelbase)(N_{LF} + N_{RF} ) + \overline{y}W \sin 15º - \overline{x}W \cos 15º = 0.[/latex]

we obtain

[latex]\overline{y} = \frac{\overline{x}W \cos 15º - (wheelbase)(N_{LF} + N_{RF} ) }{ W \sin 15º} = \frac{(1.651)(17.303) \cos 15º - (2.82)(4463 + 4396)}{17,303 \sin 15º } = 0.584 m.[/latex]

Notice that we could not have determined [latex] \overline{y}[/latex] without the measurements made with the car in the tilted position.

design Issues

The location of the center of mass of a vehicle affects its operation and performance. The forces exerted on the suspensions and wheels of cars and train coaches, the tractions their wheels create, and their dynamic behaviors are affected by the locations of their centers of mass. Not only are the performances of airplanes affected by the locations of their centers of mass, they cannot fly unless their centers of mass lie within prescribed bounds. For engineers who design vehicles, the position of the center of mass is one of the principal parameters governing decisions about the configuration of the vehicle and the layout of its contents. In testing new designs of both land vehicles and airplanes, the position of the center of mass is affected by the configuration of the particular vehicle and the weights and locations of stowage and passengers. It is often necessary to locate the center of mass experimentally by a technique such as the one we have described. Such experimental measurements are also used to confirm center of mass locations predicted by calculations made during design.

Question 7.20: Centers of Mass of Vehicles A car is placed on a platform t...

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