# 8.1: Torque

- Page ID
- 2863

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## How does a seesaw work?

It is easier to get a seesaw to move if you push on the board near the end rather than near the middle. This is due to torque - the rotational version of force.

The amount of force applied, as well as the location and the direction of the force with respect to the axis of rotation, determines the relative difficulty in causing a rotation. The longer the length of the seesaw, the easier it is to lift a heavy object.

The **lever arm** or** moment arm** is the distance between where you apply the force and the axis of rotation. The component of a force applied perpendicular to a lever arm produces a **torque**. The torque is the product of the length of the lever arm and the force applied perpendicular to the lever arm. In **Figure** above the force F is applying a torque. For our purposes, we will consider torque to be a scalar quantity possessing clockwise and counterclockwise directions. Use the seesaw simulation below to further explore the concept of torque:

Interactive Element

## The Direction of Torque

We define the direction of the torque by noting clockwise (CW) and counterclockwise (CCW) motion of an object as a result of an applied force. Whether the object actually rotates or not is unimportant. We ask how the object *would* move were it free to do so. For example, in Figure above the force F would rotate the meter stick in a counterclockwise direction. This is the same direction we turn a jar lid in order to loosen it.

Example 8.1.1

If a force F′ was applied parallel to F, but to the left of P (see Figure above), in what direction would the meter stick turn?

**Solution**

The meter stick would turn clockwise.

You may be familiar with the expression: “Righty tighty, lefty loosey.”

We define the counterclockwise direction as positive and the clockwise direction as negative. The sign of the direction is based upon the Right Hand Rule. To understand this rule, hold your right hand with your thumb pointed up and curl your fingers into a fist. Notice that the direction your fingers curl in is counterclockwise (you’re looking down). We define the upward direction in which the thumb points as positive, and the corresponding counterclockwise torque as positive. If you turn your hand such that your thumb now points down and curl your fingers into a fist, you’ll see your fingers turn clockwise. We define the downward direction in which your thumb points as negative, and the corresponding clockwise torque as negative. (For those readers who have grown up using only digital clocks, the term clockwise originated from the rotational direction that the hands of an analog clock move; counterclockwise being the reverse rotational direction.)

## Mathematical Definition of Torque

We can state the magnitude of the torque in two ways:

(1) The product of the perpendicular distance from the axis of rotation r (to the applied force) and the perpendicular component of the force Fsinθ.

(2) The product of the perpendicular distance rsinθ to the direction in which the force acts, and F.

Both (1) and (2) are equivalent to the product of r,F, and the sine of the angle between them. The symbol for torque is the Greek letter tau (τ). Thus, we can write τ=rFsinθ, where the angle θ is the angle between vectors r and F. If the angle between r and F is 90∘ then the torque is simply τ=rF. We can see by the definition of the torque that the units of torque are *m*N* though they are usually expressed as *N*m* but never as joules even though they are dimensionally equivalent.

Example 8.1.2

An engineer is trying to turn a difficult nut. Using a long wrench, he applies a 105 N force at a distance 30.0-cm from the nut shown in Figure below. What is the magnitude of the torque applied by the plumber?

**Solution**

The lever arm is 30 cm and the angle between the force and the lever arm is 90-degrees. τ=Frsinθ=(105N)(0.30m)sin90^{∘}→31.5 N⋅m

Example 8.1.3

If the minimum torque required turning the pipe in Figure above is 31.5 N·m,, could a force smaller than 105 N be used?

**Solution**

Yes. The workman could extend the length of the handle of the tool he’s using to turn the pipe. The longer the arm of the tool, the smaller the force required. You have probably experienced this phenomenon yourself. If you have ever had a problem turning a screw, increasing the thickness of the screw driver handle enables you to provide the same torque with a smaller force. The fatter the handle, the farther your hand is from the axis of rotation, and so the smaller the force needed to turn the screw. This is how a lever works. For example, the longer the lever arm on a bottle opener, the smaller the force needed to pry the bottle cap open. You may have heard the famous dictum of Archimedes: "Give me a place to stand on and with a lever I will move the world." In other words, with a long enough lever arm, even the weakest person can move a tremendous weight.

Example 8.1.4

A door of width 0.810 m requires a minimum torque of *14.47 m*N* in order to open. The door knob is positioned 5.70 cm from the left edge of the door.

(a) What minimum force is required to open the door?

(b) If the doorknob is moved to the center of the door, what is the minimum force required in opening the door? Assume that the force acts perpendicular to the plane of the door. See Figure below.

**Solution**

(a.) The force is perpendicular to the lever arm so τ=rF=14.47 m∗N. Since the doorknob is 5.7 cm from the left edge of the door, the distance from the axis of rotation is r=0.81m−0.057m=0.753m. The force required is then:

F=14.47 m∗N0.753 m=19.216→19.2 N

(b.) Since the doorknob is now in the center of the door, it is 0.81 m2=0.405 m from the axis of rotation. Therefore, the force required is 14.47 m∗N0.405 m=35.73→35.7 N

How much force would be necessary to open the door if the doorknob were placed along the axis of rotation?

Example 8.1.5

In Figure below, a force F of 95.0 N is applied to a hinged rod of length r=2.2 m. The angle between F and r is 130-degrees. Find the magnitude of the torque that the force F exerts upon the rod.

**Solution**

Once again, τ=r(Fsinθ), but this time since θ≠90∘, the equation does not reduce to τ=rF.

Therefore, τ=(2.20m)(95.0N)sin130∘=160.1 m∗N

The drawbridge in the simulation below has a cable that pulls on the bridge in order to rotate it counter-clockwise. Try to adjust the sliders so that all the forces in the bridge system are in equilibrium. Then, play around and see what conditions cause the bridge’s cable to snap:

Interactive Element

## Summary

- Torque is the component of a force applied perpendicular to a lever arm. It can have a clockwise or counter-clockwise direction.
- Torque can be calculated by taking the product of the length of a lever arm and the force applied perpendicular to a lever arm.

## Review

1. Why do we not calculate the torque using the cosine of the angle between r and F?

2. Mathematically, the definition of torque can also be expressed as τ=(rsinθ)F. In other words, we can always assume that F, rather than a component of the force, is responsible for rotation. How can this be?

## Additional Resources

Study Guide: Circular Motion Study Guide

Video: Torque - Overview

Interactive: Orbital Motion

Video: