ch12-p036 - 36. (a) With F = ma = − μk mg the magnitude...

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Unformatted text preview: 36. (a) With F = ma = − μk mg the magnitude of the deceleration is |a| = μkg = (0.40)(9.8 m/s2) = 3.92 m/s2. (b) As hinted in the problem statement, we can use Eq. 12-9, evaluating the torques about the car’s center of mass, and bearing in mind that the friction forces are acting horizontally at the bottom of the wheels; the total friction force there is fk = μkgm = 3.92m (with SI units understood – and m is the car’s mass), a vertical distance of 0.75 meter below the center of mass. Thus, torque equilibrium leads to (3.92m)(0.75) + FNr (2.4) – FNf (1.8) = 0 . Eq. 12-8 also holds (the acceleration is horizontal, not vertical), so we have FNr + FNf = mg, which we can solve simultaneously with the above torque equation. The mass is obtained from the car’s weight: m = 11000/9.8, and we obtain FNr = 3929 ≈ 4000 N. Since each involves two wheels then we have (roughly) 2.0 × 103 N on each rear wheel. (c) From the above equation, we also have FNf = 7071 ≈ 7000 N, or 3.5 × 103 N on each front wheel, as the values of the individual normal forces. (d) Eq. 6-2 directly yields (approximately) 7.9 ×102 N of friction on each rear wheel, (e) Similarly, Eq. 6-2 yields 1.4 × 103 N on each front wheel. ...
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