Hauling a Crate

A flatbed truck is carrying a crate up a 10.0° hill, as Figure 4.34a illustrates. The coefficient of static friction between the truck bed and the crate is 0.350. Find the maximum acceleration that the truck can attain before the crate begins to slip backward relative to the truck.

Figure 4.34

(a)

A crate on a truck is kept from slipping by the static frictional force . The other forces that act on the crate are its weight and the normal force .

(b)

The free-body diagram of the crate.

Reasoning The crate will not slip as long as it has the same acceleration as the truck. Therefore, a net force must act on the crate to accelerate it, and the static frictional force

contributes to this net force. Since the crate tends to slip backward, the static frictional force is directed forward, up the hill.
As the acceleration of the truck increases,

must also increase to produce a corresponding increase in the acceleration of the crate. However, the static frictional force can increase only until its maximum value

is reached, at which point the crate and truck have the maximum acceleration

. If the acceleration increases even more, the crate will slip.
To find

, we will employ Newton's second law, the definition of weight, and the relationship between the maximum static frictional force and the normal force.
Knowns and Unknowns The data for this problem are as follows:

Description	Symbol	Value
Angle of hill	θ	10.0°
Coefficient of static friction	μ_s	0.350
Unknown Variable
Maximum acceleration before crate slips	a^MAX	?

Modeling the Problem

Newton's Second Law (x direction) With the x direction chosen to be parallel to the acceleration of the truck, Newton's second law for this direction can be written as (see Equation 4.2a)

, where ΣF_x is the net force acting on the crate in the x direction and m is the crate's mass. Using the x components of the forces shown in Figure 4.34b, we find that the net force is

. Substituting this expression into Newton's second law gives

The acceleration due to gravity g and the angle θ are known, but m and

are not. We will now turn our attention to finding

The Maximum Static Frictional Force The magnitude

of the maximum static frictional force is related to the coefficient of static friction μ_s and the magnitude F_N of the normal force by Equation 4.7:


	(4.7)

This result can be substituted into Equation 1, as shown at the right. Although μ_s is known, F_N is not known. An expression for F_N will be found in Step 3, however.

Newton's Second Law (y direction) We can determine the magnitude F_N of the normal force by noting that the crate does not accelerate in the y direction (a_y = 0 m/s²). Thus, Newton's second law as given in Equation 4.2b becomes

There are two forces acting on the crate in the y direction (see Figure 4.34b): the normal force +F_N and the y component of the weight -mg cos θ (The minus sign is included because this component points along the negative y direction.) Thus, the net force is ΣF_y = +F_N - mg cos θ. Newton's second law becomes

This result for F_N can be substituted into Equation 4.7, as indicated at the right.

Solution Algebraically combining the results of the three steps, we find that

Note that the mass m of the crate is algebraically eliminated from the final result. Thus, the maximum acceleration is