Question 16.SP.1: When the forward speed of the van shown is 30 ft/s, the brak...
When the forward speed of the van shown is 30 ft/s, the brakes are suddenly applied, causing all four wheels to stop rotating. The van skids to rest in 20 ft. Determine the magnitude of the normal reaction and of the friction force at each wheel as the van skids to rest.
STRATEGY: You are given enough information to determine the acceleration and you want to find forces, so use Newton’s second law. The motion described is pure translation, so the angular acceleration is zero.
MODELING: Choose the van to be your system and model it as a rigid body. A free-body diagram and a kinetic diagram for this system are shown in Fig. 1. 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 and F_A represent the sum of the reactions at the rear wheels, while N_B and F_B represent the sum of the reactions at the front wheels. Since the truck is in translation, α = 0 and the inertial terms reduce to the vector m \overline{\mathbf{a}} attached at G.
ANALYSIS:Kinematics of Motion. Choose the positive sense to the right and use the equations of uniformly accelerated motion. You have
\begin{gathered}\bar{v}_0=+30 \mathrm{ft} / \mathrm{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 \mathrm{ft} / \mathrm{s}^2 \quad \overline{\mathbf{a}}=22.5 \mathrm{ft} / \mathrm{s}^2 \leftarrow\end{gathered}
Equations of Motion. You can obtain three equations of motion by expressing that the system of the external forces from your free-body diagram is equivalent to the inertial terms from your kinetic diagram. Applying Newton’s second law in the x and y directions gives
+\uparrow \Sigma F_y=m \bar{a}_y: \quad N_A + N_B – W=0 (1)
\stackrel{+}{\rightarrow} \Sigma F_x=m \bar{a}_x: \quad-\left(F_A + F_B\right)=-m \bar{a} (2)
Taking moments about any point gives you a third equation. For moments about point A, you find
+ \circlearrowleft \Sigma M_A=\bar{I} \alpha + m \bar{a} d_{\perp}: \quad-W(5 \mathrm{ft}) + N_B(12 \mathrm{ft})=m \bar{a}(4 \mathrm{ft}) (3)
In these three equations you have five unknowns, N_A, N_B, F_A, F_B, and \bar{a}. Since F_A=\mu_k N_A and F_B=\mu_k N_B, where \mu_k is the coefficient of kinetic friction, you have from Eq. (1)
F_A + F_B=\mu_k\left(N_A + N_B\right)=\mu_k W
Substituting into Eq. (2) and using m = W/g gives
-\mu_k W=-\frac{W}{32.2 \mathrm{ft} / \mathrm{s}^2} \bar{a}=-\frac{W}{32.2 \mathrm{ft} / \mathrm{s}^2}\left(22.5 \mathrm{ft} / \mathrm{s}^2\right)
or \mu_k=0.699. Solving Eq. (3) for N_B gives you N_B=0.640 \mathrm{~W}. Substituting this into Eq. (1), you find N_A=0.350 W. The friction forces are easily determined once you know the normal forces F_A=\mu_k N_A=(0.699)(0.350 W) = 0.245W and F_B=\mu_k N_B=(0.699)(0.650 W)=0.454 W.
Reactions at Each Wheel. Recall that the values computed here represent the sum of the reactions at the two front wheels or the two rear wheels. You obtain the magnitude of the reactions at each wheel by writing
\begin{array}{ll}N_{\text {front }}=\frac{1}{2} N_B=0.325 \mathrm{~W} &N_{\text {rear }}=\frac{1}{2} N_A=0.175 \mathrm{~W} \\F_{\text {front }}=\frac{1}{2} F_B=0.227 \mathrm{~W} & F_{\text {rear }}=\frac{1}{2} F_A=0.122 \mathrm{~W}\end{array}
REFLECT and THINK: Note that even though the angular acceleration of the van is zero, the sum of the moments about point A is not equal to zero, since from the kinetic diagram, m \bar{a} produces a moment about A. Rather than taking moments about point A, you also could have chosen to take moments about the center of mass, G. In this case, the sum of the moments would have been equal to zero. You only get three independent equations for a rigid body in plane motion: \Sigma F_x, \Sigma F_y, and one moment equation.