## Chapter 7

## Q. 7.6

## Q. 7.6

**BUCKLE UP FOR SAFETY**

**GOAL** Calculate the frictional force that causes an object to have a centripetal acceleration.

**PROBLEM** A car travels at a constant speed of 30.0 mi/h (13.4 m/s) on a level circular turn of radius 50.0 m, as shown in the bird’s-eye view in Figure 7.11a. What minimum coefficient of static friction, μ_s, between the tires and roadway will allow the car to make the circular turn without sliding?

**STRATEGY** In the car’s free-body diagram (Fig. 7.11b) the normal direction is vertical and the tangential direction is into the page (Step 2). Use Newton’s second law. The net force acting on the car in the radial direction is the force of static friction toward the center of the circular path, which causes the car to have a centripetal acceleration. Calculating the maximum static friction force requires the normal force, obtained from the normal component of the second law.

## Step-by-Step

## Verified Solution

(Steps 3, 4) Write the components of Newton’s second law. The radial component involves only the maximum static friction force, f_{s, max}:

-m \frac{v^2}{r} = -f_{s,max} = -\mu_s n

In the vertical component of the second law, the gravity force and the normal force are in equilibrium:

n-m g = 0 \quad → \quad n = mg

(Step 5) Substitute the expression for n into the first equation and solve for \mu_s:

-m \frac{v^2}{r} = -\mu_s mg

\qquad \mu_s = \frac{v^2}{rg} = \frac{(13.4 m/s)^2}{(50.0 m) (9.80 m/s^2)} = 0.366

**REMARKS** The value of \mu_s for rubber on dry concrete is very close to 1, so the car can negotiate the curve with ease. If the road were wet or icy, however, the value for \mu_s could be 0.2 or lower. Under such conditions, the radial force provided by static friction wouldn’t be great enough to keep the car on the circular path, and it would slide off on a tangent, leaving the roadway.