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The coefficient of static friction between the tires of a car and a dry road is 0.620.62. The mass of the car is 15001500 kg. What max- imum braking force is obtainable (a)(a) on a level road and (b)(b) on an 8.6°8.6\degree downgrade?

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(a)\textbf{(a)} The maximum braking force is equal to the maximum static friction force. The latter is given by Eq. 5-7:

Fbraking=fs,max=μsN=μsmg\begin{align*} F_{\text{braking}}=f_{s,max}=\mu_s N=\mu_s mg \end{align*}

where we have substituted N=mgN=mg since the road is level.

Inserting the numbers we get

 Fbraking=0.62(1500 kg)(9.81 ms2)    Fbraking=9123 N\begin{align*} &~ F_{\text{braking}}=0.62\cdot (1500\mathrm{~kg}) \left( 9.81\mathrm{~\frac{m}{s^2}} \right) \\ \implies &\boxed{F_{\text{braking}}=9123 \mathrm{~N}} \end{align*}

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