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A 13.5-kN automobile is being driven at 50 km/h through a turn of the radius of 60 m. Assuming that the forces are equally balanced among the four wheels,...

A 13.5-kN automobile is being driven at 50 km/h through a turn of the 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 (a = v2/r) where m is the vehicle’s mass, v denotes its speed, and r is the turn radius.

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Solution:

Approach :-

we are tasked with finding the resultant force the applied racial and throsh forces on the vehicle wheel bearinges.

The cornering force acting on the entire vehicle is $\frac{m v^2}{r}$,

We wis substitute specific values to obtain a numerical result.

with w denoting the automobile's weight, each wheel carries the radios,

Solution:

$ F_R=\frac{w}{4}\\ $

The thrust force curried by one wheel is given by the expression.

$ F_r=\frac{z}{4} \frac{m v^2}{r} $

where the velicle mass is,

$ m=\frac{w}{g}\\ $

Fence Components is.

$ \begin{aligned}\\ F &=\sqrt{F_R^2+F_T^2} \\\\ &=\frac{m}{4} \sqrt{g^2+\left(\frac{v^2}{r}\right)^2}\\ \end{aligned}\\ $

The vehicle's muss is

$ \begin{aligned}\\ m &=\frac{13.5 \times 10^3 \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^2} \\\\ &=1.376 \times 10^3\left(\frac{\mathrm{kg} . \mathrm{m}}{5^2}\right)\left(\frac{5^2}{\mathrm{~m}}\right) \\\\ &=1.376 \times 10^3 \mathrm{~kg} \\ \end{aligned} \\ $

the velocity is.

$ \begin{aligned}\\ V &=\left(50 \frac{\mathrm{km}}{\mathrm{h}}\right)\left(10^3 \frac{\mathrm{m}}{\mathrm{km}}\right)\left(\frac{1}\\{3600} \frac{\mathrm{h}}{\mathrm{s}}\right) \\\\ &=13.89\left(\frac{\mathrm{km}}{\mathrm{h}}\right)\left(\frac{\mathrm{m}}\\{\mathrm{km}}\right)\left(\frac{\mathrm{h}}{\mathrm{s}}\right) \\\\ &=13.89 \frac{\mathrm{m}}{\mathrm{s}}\\ \end{aligned}\\ $

The resiltunt force acting on a whee's beering is,

$ F=\left(\frac{1.376 \times 10^3 \mathrm{~kg}}{4}\right)\\ $

$ \sqrt{\left(9.81 \frac{m}{s^2}\right)^2+\left(\frac{(3.84 m / g}{60 m}\right)^2} \\ $

$ \begin{aligned} &=3551 \frac{\mathrm{k} 9 . \mathrm{m}}{\mathrm{s}^2} \\ F &=3.551 \mathrm{kN} \end{aligned} $

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