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If the 4, 500 lb van has front wheel drive and the
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Answer #1

Forces on front wheel are, N_A vertically upwards and frictional force F_A in the forward direction

Forces on rear wheel are, N_B vertically upwards and frictional force F_B=0

weight of van lies at G = W and is vertically downwards

net vertical force is zero.

so 4500=N_A+N_B ------------(1)

The maximum acceleration occurs when front wheels are about to slip.

that is F_A=\mu_s N_A

a_{max}=\frac{F_A}{m}=\frac{\mu_sN_A}{4500/32.2} -----------(2)

torques about center of gravity G ,

N_B\times6=N_A\times3.5+F_A\times2.5

N_B\times6=N_A\times3.5+\mu_s N_A\times2.5

N_B\times6=N_A\times3.5+(0.8) N_A\times2.5  ------(3)

using (1) and (3)

N_B=2152\,lb  and N_A=2347.8\,lb

substituting value of N_A=2347.8\,lb in (2)

a_{max}=\frac{(0.8)(2347.8)}{4500/32.2}=13.4\,ft/s^2

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