A 13.5-kN automobile is being driven at 50 km/h through a turn of radius of 60 m. Assuming that the forces are equally balanced among the four wheels, calculate the magnitude of the resultant force acting on the tapered roller bearings that support each wheel. To calculate the cornering force, apply

Question

Automobile Wheel Bearings

A 13.5-kN automobile is being driven at 50 km/h through a turn of radius of 60 m. Assuming that the forces are equally balanced among the four wheels, calculate the magnitude of the resultant force acting on the tapered roller bearings that support each wheel. To calculate the cornering force, apply Newton’s second law (F = ma) with the centripetal acceleration [latex](a = v^2/r)[/latex] where m is the vehicle’s mass, v denotes its speed, and r is the turn radius.

Approach
We are tasked with finding the resultant force from the applied radial and thrust forces on the vehicle wheel bearings. We assume that each wheel carries one-quarter of the vehicle’s weight, and that force component is oriented radially on the wheel’s bearings. The cornering force acting on the entire vehicle is [latex]mv^2/r[/latex], directed toward the center of the turn, and the fraction carried by each wheel is a thrust force parallel to the wheel’s axle. In terms of those variables, we will determine a general symbolic equation for the magnitude of a wheel’s resultant force, and then we will substitute specific values to obtain a numerical result. (See Figure 4.30.)

Accepted Answer

Solution
With [latex]\omega[/latex] denoting the automobile’s weight, each wheel carries the radial force

[latex]F_{R}=\frac{\omega }{4} [/latex]

The thrust force carried by one wheel is given by the expression

[latex]F_{T}=\frac{1 }{4}\frac{m\upsilon ^2}{r} \longleftarrow [F=ma][/latex]

where the vehicle’s mass is

[latex]m=\frac{\omega }{g} [/latex]

The resultant of those two perpendicular force components is

[latex]F=\sqrt{F^{2}_{R}+F^{2}_{T} } \longleftarrow [F=\sqrt{F^{2}_{x}+F^{2}_{y} } ] [/latex]

[latex]=\frac{m}{4}\sqrt{g^{2}+(\frac{\upsilon ^2}{r} )^{2} } [/latex]

Next, we will substitute the numerical values given in the problem’s statement into this general expression. The vehicle’s mass is

[latex]m=\frac{13.5 imes 10^3N}{9.81{m}/{s}^2} \longleftarrow [m=\frac{\omega }{g} ] [/latex]

[latex]=1.376 imes 10^3\left(\frac{kg.\cancel{m}}{\cancel{s^2}} ight) (\frac{\cancel{s^2}}{\cancel{m}} )[/latex]

[latex]= 1.376 × 10^3 kg[/latex]

or 1.376 Mg. In consistent dimensions, the velocity is

[latex] u = (50\frac{Km}{h} )(10^3\frac{m}{Km} )(\frac{1}{3600}\frac{h}{s} )[/latex]

[latex]=13.89(\frac{\cancel{Km}}{\cancel{h}}(\frac{m}{\cancel{Km}} )(\frac{\cancel{h}}{s} )[/latex]

[latex]=13.89\frac{m}{s}[/latex]

With a 60-m turn radius, the magnitude of the resultant force acting on a wheel’s bearings is

[latex]F=(\frac{1.376 imes 10^3Kg}{4} )\sqrt{(9.81\frac{m}{s^2} )^2+\left(\frac{(13.89m/s)^2}{60m} ight)^2 } \longleftarrow \left[F=\frac{m}{4}\sqrt{g^2+\left(\frac{\upsilon ^2}{r} ight)^2} ight] [/latex]

[latex]=3551\frac{Kg.m}{s^2}[/latex]

[latex]=355 1N[/latex]

[latex]=3.551 K N[/latex]

Discussion
Because of the cornering force, the wheel bearings carry somewhat more than one-quarter of the vehicle’s weight (3.375 kN). As a double-check, we can verify the dimensional consistency of the calculations by noting that the term [latex]v^2/r[/latex], which is combined with the gravitational acceleration g, has the dimensions of acceleration.

F = 3.551 kN

Question 4.9: A 13.5-kN automobile is being driven at 50 km/h through a tu...

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