problem11_44

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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11.44: The simplest way to do this is to consider the changes in the forces due to the extra weight of the box. Taking torques about the rear axle, the force on the front wheels is decreased by N, 1200 N 3600 m 3.00 m 1.00 = so the net force on the front wheels is N 10 9.58 N 1200 N 10,780 3 × = - to three figures. The weight added to the rear wheels is
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Unformatted text preview: then N, 4800 N 1200 N 3600 = + so the net force on the rear wheels is N, 10 36 . 1 N 4800 N 8820 4 × = + again to three figures. b) Now we want a shift of N 10,780 away from the front axle. Therefore, N 780 , 10 m 00 . 3 m 00 . 1 = W and so N. 340 , 32 = w...
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• Force, net force, rear wheels

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