6.1 The Equilibrium Model

Kinematics is a description of how an object moves. But our goal is deeper: We would like an explanation for why an object moves as it does. Galileo and Newton discovered that motion is determined by forces. In the absence of a net force, an object is at rest or moves with constant velocity. Its acceleration is zero, and this is the basis for our first explanatory model: the equilibrium model.

MODEL 6.1 Mechanical equilibrium

For objects on which the net force is zero.

Model the object as a particle with no acceleration.
- A particle at rest is in equilibrium.
- A particle moving in a straight line at constant speed is also in equilibrium.
Mathematically: $\vec{a} = \vec{0}$ in equilibrium; thus
- Newton’s second law is ${\vec{F}}_{net} = \sum_{i} {\vec{F}}_{i} = \vec{0} .$
- The forces are “read” from the free-body diagram,
Limitations: Model fails if the forces aren’t balanced.

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

Full Alternative Text

Newton’s laws are vector equations. The requirement for mechanical equilibrium, $\vec{F}_{net} = \vec{0}$ and thus $\vec{a} = \vec{0}$ , is a shorthand way of writing two simultaneous equations:

{(F_{net})}_{x} = \sum_{i} {(F_{i})}_{x} = 0 and {(F_{net})}_{y} = \sum_{i} {(F_{i})}_{y} = 0

(6.1)

In other words, the sum of all x-components and the sum of all y-components must simultaneously be zero. Although real-world situations often have forces pointing in three dimensions, thus requiring a third equation for the z-component of $\vec{F}_{net}$ , we will restrict ourselves for now to problems that can be analyzed in two dimensions.

The figure illustrates a photograph of a bridge. Note: The concept of equilibrium is essential for the engineering analysis of stationary objects such as bridges. — The concept of equilibrium is essential for the engineering analysis of stationary objects such as bridges.

NOTE The equilibrium condition of Equations 6.1 applies only to particles, which cannot rotate. Mechanical equilibrium of an extended object, which can rotate, requires an additional condition that we will study in Chapter 12.

Equilibrium problems occur frequently. Let’s look at a couple of examples.

EXAMPLE 6.1 Finding the force on the kneecap

Your kneecap (patella) is attached by a tendon to your quadriceps muscle. This tendon pulls at a $10 °$ angle relative to the femur, the bone of your upper leg. The patella is also attached to your lower leg (tibia) by a tendon that pulls parallel to the leg. To balance these forces, the end of your femur pushes outward on the patella. Bending your knee increases the tension in the tendons, and both have a tension of 60 N when the knee is bent to make a $70 °$ angle between the upper and lower leg. What force does the femur exert on the kneecap in this position?

MODEL Model the kneecap as a particle in mechanical equilibrium.

VISUALIZE FIGURE 6.1 shows how to draw a pictorial representation. We’ve chosen to align the x-axis with the femur. The three forces—shown on the free-body diagram—are labeled $\vec{T}_{1}$ and $\vec{T}_{2}$ for the tensions and $\vec{F}$ for the femur’s push. Notice that we’ve defined angle $θ$ to indicate the direction of the femur’s force on the kneecap.

The figure consists of two parts that illustrate a pictorial representation and free-body diagram of the kneecap in equilibrium.

The figure illustrates the known variables and the variable to be found.

Full Alternative Text

SOLVE This is an equilibrium problem, with three forces on the kneecap that must sum to zero. For $\vec{a} = \vec{0}$ , Newton’s second law, written in component form, is

\begin{matrix} {(F_{net})}_{x} = \sum_{i} {(F_{i})}_{x} = T_{1 x} + T_{2 x} + F_{x} = 0 \\ {(F_{net})}_{y} = \sum_{i} {(F_{i})}_{y} = T_{1 y} + T_{2 y} + F_{y} = 0 \end{matrix}

NOTE You might have been tempted to write $- T_{1 x}$ in the equation since $\vec{T}_{1}$ points to the left. But the net force, by definition, is the sum of all the individual forces. The fact that $\vec{T}_{1}$ points to the left will be taken into account when we evaluate the components.

The components of the force vectors can be evaluated directly from the free-body diagram:

\begin{array}{l} T_{1 x} = - T_{1} \cos 10^{\circ} & T_{1 y} = T_{1} \sin 10^{\circ} \\ T_{2 x} = - T_{2} \cos 70^{\circ} & T_{2 y} = - T_{2} \sin 70^{\circ} \\ F_{x} = F \cos θ & F_{y} = F \sin θ \end{array}

This is where signs enter, with $\vec{T}_{1}$ , being assigned a negative value because $\vec{T}_{1}$ points to the left. Similarly, $\vec{T}_{2}$ points both to the left and down, so both $T_{2 x}$ and $T_{2 y}$ are negative. With these components, Newton’s second law becomes

\begin{array}{l} - T_{1} cos 10 ° - T_{2} cos 70 ° + F cos θ = 0 \\ T_{1} sin 10 ° - T_{2} sin 70 ° + F sin θ = 0 \end{array}

These are two simultaneous equations for the two unknowns $F$ and $θ$ . We will encounter equations of this form on many occasions, so make a note of the method of solution. First, rewrite the two equations as

\begin{array}{l} F cos θ = T_{1} cos 10 ° + T_{2} cos 70 ° \\ F sin θ = - T_{1} sin 10 ° + T_{2} sin 70 ° \end{array}

Next, divide the second equation by the first to eliminate $F :$

\frac{F sin θ}{F cos θ} = tan θ = \frac{- T_{1} sin 10 ° + T_{2} sin 70 °}{T_{1} cos 10 ° + T_{2} cos 70 °}

Then solve for $θ :$

\begin{array}{l} θ & = & {tan}^{- 1} (\frac{- T_{1} sin 10 ° + T_{2} sin 70 °}{T_{1} cos 10 ° + T_{2} cos 70 °}) \\ = & {tan}^{- 1} (\frac{- (60 N) sin 10 ° + (60 N) sin 70 °}{(60 N) cos 10 ° + (60 N) cos 70 °}) = 30 ° \end{array}

Finally, use $θ$ to find $F :$

\begin{matrix} F & = & \frac{T_{1} cos 10 ° + T_{2} cos 70 °}{c o s θ} \\ = & \frac{(60 N) c o s 10 ° + (60 N) c o s 70 °}{cos 30 °} = 92 N \end{matrix}

The question asked What force? and force is a vector, so we must specify both the magnitude and the direction. With the knee in this position, the femur exerts a force $\vec{F} = (92 N, 30 ° above the femur)$ on the kneecap.

REVIEW The magnitude of the force would be 0 N if the leg were straight, 120 N if the knee could be bent $180 °$ so that the two tendons pull in parallel. The knee is closer to fully bent than to straight, so we would expect a femur force between 60 N and 120 N. Thus the calculated magnitude of 92 N seems reasonable.

EXAMPLE 6.2 Towing a car up a hill

A car with a weight of 15,000 N is being towed up a $20 °$ slope at constant velocity. Friction is negligible. The tow rope is rated at 6000 N maximum tension. Will it break?

MODEL Model the car as a particle in mechanical equilibrium.

VISUALIZE Part of our analysis of the problem statement is to determine which quantity or quantities allow us to answer the yes-or-no question. In this case, we need to calculate the tension in the rope. FIGURE 6.2 shows the pictorial representation. Note the similarities to Examples 5.2 and 5.6 in Chapter 5, which you may want to review.

The figure illustrates a free-body diagram of a car being towed up a hill.

Full Alternative Text

We noted in Chapter 5 that the weight of an object at rest is the magnitude $F_{G}$ of the gravitational force acting on it, and that information has been listed as known.

SOLVE The free-body diagram shows forces $\vec{T}$ $\vec{n}$ , and $\vec{F}_{G}$ acting on the car. Newton’s second law with $\vec{a} = \vec{0}$ is

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

From here on, we’ll use $\sum F_{x}$ and $\sum F_{y}$ , without the label $i$ , as a simple shorthand notation to indicate that we’re adding all the x-components and all the y-components of the forces.

We can find the components directly from the free-body diagram:

\begin{array}{l} T_{x} = T & T_{y} = 0 \\ n_{x} = 0 & n_{y} = n \\ {(F_{G})}_{x} = - F_{G} s i n θ & {(F_{G})}_{y} = - F_{G} c o s θ \end{array}

NOTE The gravitational force has both x- and y-components in this coordinate system, both of which are negative due to the direction of the vector $\vec{F}_{G}$ . You’ll see this situation often, so be sure you understand where $(F_{G})_{x}$ and $(F_{G})_{y}$ come from.

With these components, the second law becomes

\begin{array}{l} T - F_{G} s i n θ = 0 \\ n - F_{G} c o s θ = 0 \end{array}

The first of these can be rewritten as

T = F_{G} s i n θ = (15, 000 N) s i n 20 ° = 5100 N

Because $T < 6000 N,$ we conclude that the rope will not break. It turned out that we did not need the y-component equation in this problem.

REVIEW Because there’s no friction, it would not take any tension force to keep the car rolling along a horizontal surface $(θ = 0 °)$ . At the other extreme, $θ = 90 °$ , the tension force would need to equal the car’s weight $(T = 15, 000 N)$ to lift the car straight up at constant velocity. The tension force for a $20 °$ slope should be somewhere in between, and 5100 N is a little less than half the weight of the car. That our result is reasonable doesn’t prove it’s right, but we have at least ruled out careless errors that give unreasonable results.

6.1 The Equilibrium Model

MODEL 6.1 Mechanical equilibrium

EXAMPLE 6.1 Finding the force on the kneecap

FIGURE 6.1 Pictorial representation of the kneecap in equilibrium.

EXAMPLE 6.2 Towing a car up a hill

FIGURE 6.2 Pictorial representation of a car being towed up a hill.