An automobile is braking on a flat, dry road with a coefficient of static friction of 0.90 between its wheels and the road. The wheelbase (the distance between the front and the rear wheels) is 3.0 m, and the center of mass is midway between the wheels, at a height of 0.60 m above the road.
What is the deceleration if the rear-wheel brakes are disabled? Take into account that during braking, the normal force on the front wheels is larger than that on the rear wheels.
I was trying:
$$\Sigma F_X = f_r+f_f = ma$$ $$\Sigma f_Y = N_f+N_r=mg$$ $$\Sigma \tau = ma(0.6)+N_F(1.5)+N_r(1.5)+f_r(1.5)+f_f(1.5)$$
where the sub f and sub r represent the rear and the front wheels components of normal and friction respectively
solving for a in the 3 equation
$$ma(0.6) = 1.5(mg)+1.5(ma)$$ $$a = \frac{1.5}{2.1}g$$
however this answer IS WRONG can someone tell me what I get wrong?
Note that the rear brakes are disabled. So the friction force on the rear is zero. Then $$f_f=0.9N_f=ma\\N_r+N_f=mg$$ For the torque equation, write it around the point of contact of the front wheel with the road (the torque of $N_f$ and $f_f$ are zero): $$3N_r+0.6ma-1.5mg=0$$ Taking $N_f=ma/0.9$from the first equation, we get $$N_r=mg-\frac{ma}{0.9}$$ Plug this into the torque equation: $$3mg-\frac{ma}{0.3}+0.6ma-1.5mg=0\\1.5mg=ma\left(\frac1{0.3}-0.6\right)\\1.5g=a2.733\\a=\frac{1.5}{2.733}g=0.548g=0.548\cdot 9.8m/s^2=5.37m/s^2$$