# Chapter 11: Torque

## 11.5 Equilibrium revisited

An object is in static equilibrium when the following conditions are true:

1. The net torque acting on the object is zero: $$\tau_\textit{net} = 0$$
2. The net force acting on the object is zero: $$\vec{F}_\textit{net} = 0$$ Remember that force is a vector, so both both force components ($$F_x$$ and $$F_y$$) are zero.

When an object is in static equilibrium, and not rotating at all, we are free to choose a convenient reference for calculating torques. This of course raises the question, “what makes a convenient choice?”

Often, you won’t immediately know the magnitude of a particular force exerted on an object. If the object is in static equilibrium, you can use the point where this force acts as the reference point for torque calculations, totally bypassing the problem of not knowing anything about that force! How convenient!

#### example 11.6

A 2 m long rod with a uniform mass distribution is supported 40 cm from the left end. When you apply a 1.5 N force directly down on the left end, the rod will rest in a horizontal position. What is the mass of the rod?

##### E-FBD—Rod

Since the rod is remaining in a horizontal position, it has no angular acceleration; it is in static equilibrium. I don’t know the normal force of the support on the rod, so I’ll use this location as the reference point for torque calculations. All the forces are acting perpendicular to the rod. Since $$\sin(90^\circ) = 1$$, I’ll leave the $$\sin\theta$$ factor out of the expressions of torque.

##### Newton’s second law (for rotation)

\begin{align*} \tau_\textit{net} &= I\alpha \\ \tau^\textit{app} – \tau^G &= I(0) \\ F^\textit{app}r_\textit{app} – F^Gr_\textit{cm} &= 0 \\ F^\textit{app}r_\textit{app} – mgr_\textit{cm} &= 0 \\ \hookrightarrow m &= \frac{F^\textit{app}r_\textit{app}}{gr_\textit{cm}} \end{align*}

The center of mass is in the center of the rod, 1 m from the left end—this is 0.6 m from the reference I’m using for torque calculations. We then have

\begin{align*} m &= \frac{(1.5\ \textrm{N})(0.40\ \textrm{m})}{\left(9.81\ \textrm{m/s}^2\right)(0.60\ \textrm{m})} \\ &= 0.102\ \textrm{kg} \end{align*}

#### Example 11.7

The distance from the elbow a Tyrannosaurus Rex’s elbow to the tip of its fingers is about 90 cm. The T. Rex bicep connected to the forearm about 4 cm from the elbow. When holding its arm in a horizontal position, what is the force that be bicep exerts on the forearm? What is the force exerted on the forearm by the elbow joint?

We’ll use a simple model where we say that a T. Rex arm has a uniformly distributed mass of 93 kg, and where the bicep pulls perpendicular to the forearm when held in this position.

##### E-FBD—Forearm

The elbow joint of course not just a single contact point between two bones; there are multiple bones, tendons, and ligaments involved. However, we can simplify the interaction between the elbow and the forearm as a single net force directed at some arbitrary angle. We know neither the magnitude nor direction of this force—so we choose this point as the reference for torque calculations! With this choice, the elbow exerts no torque on the forearm.

We begin by analyzing torque to determine the tension in the bicep.

##### Newton’s second law (for rotation)

\begin{align*} \tau_\textit{net} &= I\alpha \\ \tau^T – \tau^G &= I(0) \\ F^Tr_T – F^Gr_\textit{cm} &= 0 \\ F^Tr_T – mgr_\textit{cm} &= 0 \\ \hookrightarrow F^T &= \frac{mgr_\textit{cm}}{r_T} \\ &= \frac{(93\ \textrm{kg})\left(9.81\ \textrm{m/s}^2\right)(0.45\ \textrm{m})}{0.04\ \textrm{m}} \\ &= 10.26\ \textrm{kN} \end{align*}

Now that we know the force applied by the bicep, we can apply Newton’s second law using force, rather than torque, to determine the force exerted by the elbow on the forearm.

##### Newton’s second law

\begin{align*} F_\textit{net,x} &= ma_x & F_\textit{net,y} &= ma_y \\ F^\textit{elbow}_x &= m(0) & F^\textit{elbow}_y + F^T – F^G &= m(0) \\ &= 0 & F^\textit{elbow}_y + F^T – mg &= 0 \\ & & F^\textit{elbow}_y &= mg – F^T \\ & & &= (93\ \textrm{kg})\left(9.81\ \textrm{m/s}^2\right) – (10.26 \times 10^3\ \textrm{N}) \\ & & &= -9.35\ \textrm{kN} \end{align*}

There are two things I would like to point out about the force the elbow exerts on the forearm:

1. The elbow appears to exert no force in the horizontal direction. This is because of our simplification that the bicep is pulling directly upwards—nothing is acting in the horizontal direction.
2. The force the elbow exerts came out negative; we have inherently defined “up” as the positive direction (by saying $$F^T$$ was in the $$+y$$ direction), so the elbow is pushing down on the forearm.

#### Example 11.8

A cat with a mass of 5 kg is standing at the end of a horizontal boom of length $$L$$. A cable is attached $$\frac{2}{3}$$ of the way out, at an angle of $$\theta = 45^\circ$$ with the boom. The boom is attached to the wall by a hinge. The mass of the boom is 100 kg, uniformly distributed. What is the tension in the cable? What is the magnitude of the force that the pin in the hinge exerts on the boom?

##### E-FBD—Boom

As with the previous example, we know nothing about the force the pin exerts on the boom. (However, unlike the previous example, we can at least identify this force as a normal force.) This means we should choose the hinge as the reference point to calculate torques.

##### Newton’s second law (for rotation)

\begin{align*} \tau_\textit{net} &= I\alpha \\ \tau^T – \tau^G_b – \tau^G_c &= I(0) \\ F^Tr_T\sin\theta – m_bgr_\textit{cm} – m_cgr_c &= 0 \\ F^T\frac{2L}{3}\sin\theta – m_bg\frac{L}{2} – m_cgL &= 0 \end{align*}

Note that the total length of the boom $$L$$ divides out.

$\hookrightarrow F^T = \frac{\left(\frac{m_b}{2} + m_c\right)g}{\frac{2}{3}\sin\theta}$

It is standard practice to simplify all fractions. Remove the fraction from the denominator by multiplying both numerator and denominator by 3. Remove the fraction from the numerator my multiplying both numerator and denominator by 2.

\begin{align*} F^T &= \frac{3\left(\frac{m_b}{2} + m_c\right)g}{2\sin\theta} \\ &= \frac{3\left(m_b + 2m_c\right)g}{4\sin\theta} \\ &= \frac{3\left((100\ \textrm{kg}) + 2(5\ \textrm{kg})\right)\left(9.81\ \textrm{m/s}^2\right)}{4\sin\left(45^\circ\right)} \\ &= 1.14\ \textrm{kN} \end{align*}

And now that we know the tension in the cable, we can find the force exerted by the hinge.

##### Newton’s second law

\begin{align*} F_\textit{net,x} &= m_ba_x & F_\textit{net,y} &= m_ba_y \\ F^N_x – F^T_x &= 0 & F^N_y + F^T_y – F^G_b – F^G_c &= 0 \\ \hookrightarrow F^N_x &= F^T\cos\theta & \hookrightarrow F^N_y &= (m_b + m_c)g – F^T\sin\theta \\ &= \left(1.14 \times 10^3\ \textrm{N}\right)\cos\left(45^\circ\right) & &= ((100\ \textrm{kg}) + (5\ \textrm{kg}))\left(9.81\ \textrm{m/s}^2\right) – \left(1.14 \times 10^3\ \textrm{N}\right)\sin\left(45^\circ\right) \\ &= 806\ \textrm{N} & &= 224\ \textrm{N} \\ \end{align*}

To find the magnitude of this force:

$F^N = \sqrt{\left(F^N_x\right)^2 + \left(F^N_y\right)^2} = 837\ \textrm{N}$