6.2 Using Newton’s Second Law

The essence of Newtonian mechanics, introduced in 《 SECTION 5.4, can be expressed in two steps:

The forces acting on an object determine its acceleration $\vec{a} = \vec{F}_{net} / m$ .
The object’s trajectory can be determined by using $\vec{a}$ in the equations of kinematics.

These two ideas are the basis of a problem-solving strategy.

PROBLEM-SOLVING STRATEGY 6.1
Newtonian mechanics

MODEL Model the object as a particle. Make other simplifications depending on what kinds of forces are acting.

VISUALIZE Draw a pictorial representation.

Show important points in the motion with a sketch, establish a coordinate system, define symbols, and identify what the problem is trying to find.
Use a motion diagram to determine the object’s acceleration vector $\vec{a}$ . The acceleration is zero for an object in equilibrium.
Identify all forces acting on the object at this instant and show them on a free-body diagram.
It’s OK to go back and forth between these steps as you visualize the situation.

SOLVE The mathematical representation is based on Newton’s second law:

{\vec{F}}_{net} = \sum_{i} {\vec{F}}_{i} = m \vec{a}

The forces are “read” directly from the free-body diagram. Depending on the problem, either

Solve for the acceleration, then use kinematics to find velocities and positions; or
Use kinematics to determine the acceleration, then solve for unknown forces.

REVIEW Check that your result has correct units and significant figures, is reasonable, and answers the question.

Workbook Exercises 23

Newton’s second law is a vector equation. To apply the step labeled Solve, you must write the second law as two simultaneous equations:

\begin{array}{l} {(F_{net})}_{x} = \sum^{​} F_{x} = m a_{x} \\ {(F_{net})}_{y} = \sum^{​} F_{y} = m a_{y} \end{array}

(6.2)

The primary goal of this chapter is to illustrate the use of this strategy.

EXAMPLE 6.3 Speed of a towed car

A 1500 kg car is pulled by a tow truck. The tension in the tow rope is 2500 N, and a 200 N friction force opposes the motion. If the car starts from rest, what is its speed after 5.0 seconds?

MODEL Model the car as an accelerating particle. We’ll assume, as part of our interpretation of the problem, that the road is horizontal and that the direction of motion is to the right.

VISUALIZE FIGURE 6.3 shows the pictorial representation. We’ve established a coordinate system and defined symbols to represent kinematic quantities. We’ve identified the speed $v_{1}$ , rather than the velocity $v_{1 x}$ , as what we’re trying to find.

The figure titled Sketch illustrates a pictorial representation of a car being towed.

The figure titled Motion diagram and forces illustrates two parts that depict a motion diagram and free body diagram of a car being towed.

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SOLVE We begin with Newton’s second law:

\begin{array}{l} {(F_{net})}_{x} = \sum^{​} F_{x} = T_{x} + f_{x} + n_{x} + {(F_{G})}_{x} = m a_{x} \\ {(F_{net})}_{y} = \sum^{​} F_{y} = T_{y} + f_{y} + n_{y} + {(F_{G})}_{y} = m a_{y} \end{array}

All four forces acting on the car have been included in the vector sum. The equations are perfectly general, with $+$ signs everywhere, because the four vectors are added to give $\vec{F}_{net}$ . We can now “read” the vector components from the free-body diagram:

\begin{array}{l} T_{x} = + T & T_{y} = 0 & n_{x} = 0 & n_{y} = + n \\ f_{x} = - f & f_{y} = 0 & {(F_{G})}_{x} = 0 & {(F_{G})}_{y} = - F_{G} \end{array}

The signs, which we had to insert by hand, depend on which way the vectors point. Substituting these into the second-law equations and dividing by m give

\begin{matrix} a_{x} & = & \frac{1}{m} (T - f) \\ = & \frac{1}{1500 kg} (2500 N - 200 N) = 1.53 m / s^{2} \\ a_{y} & = & \frac{1}{m} (n - F_{G}) \end{matrix}

NOTE Newton’s second law has allowed us to determine $a_{x}$ exactly but has given only an algebraic expression for $a_{y}$ . However, we know from the motion diagram that $a_{y} = 0!$ That is, the motion is purely along the x-axis, so there is no acceleration along the y-axis. The requirement $a_{y} = 0$ allows us to conclude that $n = F_{G}$ .

Because $a_{x}$ is a constant $1.53 m / s^{2}$ , we can finish by using constant-acceleration kinematics to find the velocity:

\begin{matrix} v_{1 x} & = & v_{0 x} + a_{x} Δ t \\ = & 0 + (1.53 m / s^{2}) (5.0 s) = 7.7 m / s \end{matrix}

The problem asked for the speed after 5.0 s, which is $v_{1} =$ $7.7 m / s$ .

REVIEW $7.7 m / s \approx 15 mph$ , a quite reasonable speed after 5 s of acceleration.

If all the forces acting on an object are constant, as in the last example, then the object moves with constant acceleration and we can deploy the constant-acceleration model of kinetics. Now not all forces are constant—you will later meet forces that vary with position or time—but in many situations it is reasonable to model the motion as being due to constant forces. The constant-force model will be our most important dynamics model for the next several chapters.

MODEL 6.2 Constant force

For objects on which the net force is constant.

Model the object as a particle with uniform acceleration.
- The particle accelerates in the direction of the net force.
Mathematically:
- Newton’s second law is ${\vec{F}}_{net} = \sum_{i} {\vec{F}}_{i} = m \vec{a}$ .
- Use the kinematics of constant acceleration.
Limitations: Model fails if the forces aren’t constant.

The figure illustrates the free body diagram of an object depicted as a dot and explains the model for constant force.

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Example 6.4 Altitude of a rocket

A 500 g model rocket with a weight of 4.90 N is launched straight up. The small rocket motor burns for 5.00 s and has a steady thrust of 20.0 N. What maximum altitude does the rocket reach?

MODEL We’ll model the rocket as a particle acted on by constant forces by neglecting the velocity-dependent air resistance (rockets have very aerodynamic shapes) and neglecting the mass loss of the burned fuel.

VISUALIZE The pictorial representation of FIGURE 6.4 finds that this is a two-part problem. First, the rocket accelerates straight up. Second, after exhausting its fuel, the rocket continues going up as it slows down, a free-fall situation. The maximum altitude is at the end of the second part of the motion.

The figure illustrates a pictorial representation of a rocket launch.

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The figure illustrates two parts that depict a motion diagram and free body diagram of a rocket launch.

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SOLVE We now know what the problem is asking, have established relevant symbols and coordinates, and know what the forces are. We begin the mathematical representation by writing Newton’s second law, in component form, as the rocket accelerates upward. The free-body diagram shows two forces, so

\begin{array}{l} {(F_{net})}_{x} = \sum^{​} F_{x} = {(F_{thrust})}_{x} + {(F_{G})}_{x} = m a_{0 x} \\ {(F_{net})}_{y} = \sum^{​} F_{y} = {(F_{thrust})}_{y} + {(F_{G})}_{y} = m a_{0 y} \end{array}

The fact that vector $\vec{F}_{G}$ points downward—and which might have tempted you to use a minus sign in the y-equation—will be taken into account when we evaluate the components. None of the vectors in this problem has an x-component, so only the y-component of the second law is needed. We can use the free-body diagram to see that

\begin{array}{l} {(F_{thrust})}_{y} = + F_{thrust} \\ {(F_{G})}_{y} = - F_{G} \end{array}

This is the point at which the directional information about the force vectors enters. The y-component of the second law is then

\begin{matrix} a_{0 y} & = & \frac{1}{m} (F_{thrust} - F_{G}) \\ = & \frac{20.0 N - 4.90 N}{0.500 kg} = 30.2 m / s^{2} \end{matrix}

Notice that we converted the mass to SI units of kilograms before doing any calculations and that, because of the definition of the newton, the division of newtons by kilograms automatically gives the correct SI units of acceleration.

The acceleration of the rocket is constant until it runs out of fuel, so we can use constant-acceleration kinematics to find the altitude and velocity at burnout $(Δ t = t_{1} = 5.00 s) :$

\begin{matrix} y_{1} & = & y_{0} + v_{0 y} Δ t + \frac{1}{2} a_{0 y} {(Δ t)}^{2} \\ = & \frac{1}{2} a_{0 y} {(Δ t)}^{2} = 377 m \\ v_{1 y} & = & v_{0 y} + a_{0 y} Δ t = a_{0 y} Δ t = 151 m / s \end{matrix}

The only force on the rocket after burnout is gravity, so the second part of the motion is free fall. We do not know how long it takes to reach the top, but we do know that the final velocity is $v_{2 y} = 0$ . Constant-acceleration kinematics with $a_{1 y} = - g$ gives

v_{2 y}^{2} = 0 = v_{1 y}^{2} - 2 g Δ y = v_{1 y}^{2} - 2 g (y_{2} - y_{1})

which we can solve to find

\begin{array}{l} y_{2} & = y_{1} + \frac{v_{1 y}^{2}}{2 g} = 377 m + \frac{{(151 m/s)}^{2}}{2 (9.80 {m/s}^{2})} \\ = 1540 m = 1 .54 km \end{array}

REVIEW The maximum altitude reached by this rocket is 1.54 km, or just slightly under one mile. While this does not seem unreasonable for a high-acceleration rocket, the neglect of air resistance was probably not a terribly realistic assumption.

6.2 Using Newton’s Second Law

PROBLEM-SOLVING STRATEGY 6.1
Newtonian mechanics

EXAMPLE 6.3 Speed of a towed car

FIGURE 6.3 Pictorial representation of a car being towed.

MODEL 6.2 Constant force

Example 6.4 Altitude of a rocket

FIGURE 6.4 Pictorial representation of a rocket launch.