Question

The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and the tires is μs = 0.41, determine the greatest slope θ the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder at constant velocity.

Answer 1

Answer:

$\theta = 22.29$

Explanation:

Taking summation of force at perpendicular  to the plane

$\sum F_p = 0$

$2N_A +2N_B -mgcos\theta = 0$

$N_A +N_B - mgcos\theta = 0

$

$N_A +N_B = mgcos\theta$

Taking summation along the plane, therefore we have

$\sum Fa = 0$

$f_A +f_B -mgsin\theta = 0$  

$\mu N_A+\mu N_B - mgsin\theta = 0$

$\mu(N_A +N_B) = mgcos\theta$

from equation 1 and 2 we have

$\mu =\frac{sin\theta}{cos\theta}$

$\mu = 0.41$

$\mu = tan\theta

$

$\theta = tan^{-}\mu$

$\theta = 22.29$

Answer 2

The greatest slope the shoulder can have without causing the car to slip or tip over is 22.3⁰.

The given parameters;

• mass of the car, m = 1.6 Mg
• coefficient of static friction, $\mu_ s$ = 0.41

The greatest slope of the shoulder is determined by taking net force on the car as shown below;

$\Sigma F= ma \\\\Wsin\theta - \mu_s Wcos \theta = ma$

at constant velocity, acceleration, a, = 0

$Wsin\theta - \mu_s W cos \theta = 0\\\\W sin\theta = \mu_s Wcos \theta \\\\\mu_s = \frac{Wsin\theta }{W cos \theta } \\\\\mu_s = tan \theta \\\\\theta = tan^{-1}(\mu_s)\\\\\theta = tan^{-1}(0.41)\\\\\theta =22.3\ ^0$

Thus, the greatest slope the shoulder can have without causing the car to slip or tip over is 22.3⁰.

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