Question 16.SP.1: When the forward speed of the truck shown was 30 ft/s, the b......

Kinematics of Motion. Choosing the positive sense to the right and using the equations of uniformly accelerated motion, we write

\begin{gathered}\bar{v}_0=+30\text{ ft/s}\quad \bar{v}^2=\bar{v}_0^2+2 \bar{a} \bar{x} \quad 0=(30)^2+2\bar{a}(20) \\\bar{a}=-22.5\text{ ft/s}^2 \quad \overline{ \mathbf{a} }=22.5\text{ ft/s}^2 \leftarrow\end{gathered}

Equations of Motion. The external forces consist of the weight W of the truck and of the normal reactions and friction forces at the wheels. (The vectors N _A \text { and } F _A represent the sum of the reactions at the rear wheels, while N _B \text { and } F _B represent the sum of the reactions at the front wheels.) Since the truck is in translation, the effective forces reduce to the vector m\overline{\mathbf{a} } attached at G. Three equations of motion are obtained by expressing that the system of the external forces is equivalent to the system of the effective forces.

Since F_A=\mu_k N_A \text { and } F_B=\mu_k N_B \text {, where } \mu_k is the coefficient of kinetic friction, we find that

F_A+F_B=\mu_k\left(N_A+N_B\right)=\mu_k W
\begin{aligned}\stackrel{+}{\rightarrow} \Sigma F_x=\Sigma\left(F_x\right)_{\text {eff }}: \quad-\left(F_A+F_B\right) & =-m \bar{a} \\-\mu_k W & =-\frac{W}{32.2\text{ ft/s}^2}\left(22.5\text{ ft/s}^2\right) \\\mu_k & =0.699\end{aligned}
\begin{aligned}+\uparrow \Sigma M_A=\Sigma\left(M_A\right)_{\text{ eff}}: \quad-W(5\text{ ft}) & +N_B(12\text{ ft} )=m \overline{ a }(4 \text{ ft} ) \\-W(5 \text{ ft} )+N_B(12 \text{ ft} ) & =\frac{W}{32.2 \text{ ft} / s ^2}\left(22.5 \text{ ft} / s^2\right)(4 \text{ ft} ) \\N_B & =0.650 W\end{aligned}
F_B=\mu_k N_B=(0.699)(0.650 W ) \quad F_B=0.454 W
\begin{gathered}+\uparrow \Sigma F_y=\Sigma\left(F_y\right)_{\text {eff }}: \quad N_A+N_B-W=0 \\N_A+0.650 W-W=0 \\N_A=0.350 W\end{gathered}
F_A=\mu_k N_A=(0.699)(0.350 W ) \quad F_A=0.245 W

Reactions at Each Wheel. Recalling that the values computed above represent the sum of the reactions at the two front wheels or the two rear wheels, we obtain the magnitude of the reactions at each wheel by writing

\begin{array}{rll}N_{\text {front }}=\frac{1}{2} N_B=0.325 W & N_{\text {rear }}=\frac{1}{2} N_A=0.175 W\end{array}

\begin{array}{rll}F_{\text {front }}=\frac{1}{2} F_B=0.227 W & F_{\text {rear }}=\frac{1}{2} F_A=0.122 W\end{array}