Consider again the problem of a tipping car in Example 8.6. This time, instead of applying...

Consider again the problem of a tipping car in Example 8.6. This time, instead of applying a force F to the car, assume the car is traveling around a curve on a level road. Let the radius of curvature of the turn be r and assume the car’s speed v is just barely fast enough to make the car’s inside wheels lift off the ground, (a) Find v. Express your answer in terms of r and the variables w, d, and h from Figure 8.25. (b) Use realistic values to get a numerical estimate for v for r = 80 m. Hint: Use the car’s center of mass as your pivot point.

Figure 8.25 Example 8.6. If a force is applied horizontally to a car, the car may tip, that is, rotate about the point P.

Example 8.6

Cars and Static Equilibrium

Figure 8.25 shows the forces acting on a car in a collision. The force comes from another car, and for simplicity we assume this force is directed horizontally and acts at the center of the body of the car. Estimate the magnitude of for which the car in Figure 8.25 will just roll over. Express the answer as a ratio where m is the mass of the car, and use the dimensions of a typical car in your calculation.

RECOGNIZE THE PRINCIPLE

We wish to calculate the minimum force that will just barely cause the car to tip. This problem is very similar to the tipping crate in Figure 8.24. When the car tips, it will rotate about point P in Figure 8.25, so we take that as the pivot point.

Figure 8.24 A Forces on a crate. As the person pushes harder and harder, the crate will eventually tip and B rotate about a pivot point P at the corner of the crate.

SKETCH THE PROBLEM

Figure 8.25 shows all the forces on the car, including the force of friction between the tires and road (Ffriction) and the force of gravity (Fgrav = mg). It also shows the points at which these forces act.

IDENTIFY THE RELATIONSHIPS

Although this problem is similar to the tipping crate in Figure 8.24, there are some important differences. Because the car sits on four wheels, we must account for the height of the wheels when finding the car’s center of mass. Indeed, we are not given any information about the dimensions of the car’s body (w and h in Fig. 8.25) or the wheels (d), so we’ll also have to estimate their values. First, though, let’s express the conditions for equilibrium in terms of w, h, and d. We will make estimates for their values below.

We treat the car as a simple “box” of width w and height h that is a distance d off the ground and ignore the mass in the wheels. As with the tipping crate, there are four forces on the car as indicated in Figure 8.25. The condition for rotational equilibrium is

where the lever arm factors are given in parentheses. The term involving Fcollision is the torque produced by the collision force. The lever arm in this case is the distance from the center of the car to the ground d + h/2 since this force acts at the middle of the car, at its center of mass. This force is directed to the right (along + x), and it would, if it acted alone, produce a clockwise rotation; hence, this torque is negative.

SOLVE

We can rearrange Equation (1) to get

Dividing Fcollision by the weight of the car leads to

What does it mean?

If Fcollision/mg is small, a (relatively) small force is enough to cause a rollover accident, whereas a large ratio means that a larger impact force is needed to cause a rollover. Equation (1) shows that a large value of d—that is, a large ground clearance—will make a rollover more likely. For a typical compact car that sits relatively low to the ground, we estimate2 w = 170 cm, d = 20 cm, and h = 120 cm, which gives the ratio

For an SUV, we estimate w = 170 cm, d = 40 cm, and h = 140 cm, which leads to

Hence, according to this measure, the SUV is more likely to roll over. Of course, real accidents are more complicated than this simple model. Current government rollover ratings are based in part on a similar calculation.