Question 4.SP.14: An 8400-kg truck is traveling on a level horizontal curve, r...
An 8400-kg truck is traveling on a level horizontal curve, resulting in an effective lateral force H (applied at the center of gravity G of the truck). Treating the truck as a rigid system with the center of gravity shown, and knowing that the distance between the outer edges of the tires is 1.8 m, determine (a) the maximum force H before tipping of the truck occurs, (b) the minimum coefficient of static friction between the tires and roadway such that slipping does not occur before tipping.
STRATEGY: For the direction of H shown, the truck would tip about the outer edge of the right tire. At the verge of tip, the normal force and friction force are zero at the left tire, and the normal force at the right tire is at the outer edge. You can apply equilibrium to determine the value of H necessary for tip and the required friction force such that slipping does not occur.
MODELING: Draw the free-body diagram of the truck (Fig. 1), which reflects impending tip about point B. Obtain the weight of the truck by multiplying its mass of 8400 kg by g = 9.81 m/s²; that is, W = 82 400 N or 82.4 kN.
ANALYSIS:
Free Body: Truck (Fig. 1).
\begin{array}{ll}+\circlearrowleft \Sigma M_B=0: & (82.4 kN)(0.8 m)-H(1.4 m)=0 \\& H=+47.1 kN \\ H = 47.1 kN \rightarrow \\\underrightarrow{+}\Sigma F_x=0: \quad & 47.1 kN-F_B=0 \\ & F_B=+47.1 kN \\+\uparrow \Sigma F_y=0: & N_B-82.4 kN=0 \\& N_B=+82.4 kN\end{array}Minimun Coefficient of Static Friction. The magnitude of the maximum friction force that can be developed is
F_m=\mu_s N_B=\mu_s(82.4 kN)Setting this equal to the friction force required, F_B=47.1 kN, gives
\mu_s(82.4 kN)=47.1 kN\mu_s=0.572
REFLECT and THINK: Recall from physics that H represents the force due to the centripetal acceleration of the truck (of mass m), and its magnitude is
H = m(v²/ρ)
where
v = velocity of the truck
ρ = radius of curvature
In this problem, if the truck was traveling around a curve of 100-m radius (measured to G), the velocity at which it would begin to tip would be 23.7 m/s (or 85.2 km/h). You will learn more about this aspect in the study of dynamics.
