p13_008.pdf

1 . 78 = 1 . 27m is the horizontal distance from the rear axle to the

center of mass; F 1 is the force exerted on each front wheel; and, F 2 is the force exerted on each back

wheel.

−

(a) Taking torques about the rear axle, we ﬁnd

F 1 = Mg

2 L

= (1360kg) 9 . 8m / s 2 (1 . 27m)

2(3 . 05m)

=2 . 77

10 3 N .

(b) Equilibrium of forces leads to 2 F 1 +2 F 2 = Mg ,fromwhichweobtain F 2 =3 . 89

10 3 N.

8. Our notation is as follows: M = 1360kg is the mass of the automobile; L =3 . 05m is the horizontal

distance between the axles; =3 . 05