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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. 129, 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. 128 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. 62 directly yields (approximately) 7.9 ×102 N of friction on each rear wheel, (e) Similarly, Eq. 62 yields 1.4 × 103 N on each front wheel. ...
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 Spring '08
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 Physics, Center Of Mass, Mass

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