**Twisting** is a unique kind of force. Rather than result in linear acceleration of an object, **torque** (pronounced *"tork"*) results in rotational motion of that object about one or more axes that pass through its center of mass. One thing you should keep in mind is that any object rotating freely, like a diver doing somersaults, rotates around a line passing through its center of mass.

In this section we'll develop the idea of torque conceptually and mathematically – as we do for any physical quantity.

You are already familiar with the idea of torque if you've ever used a screwdriver or a wrench to turn a screw or bolt, or if you've used a lever to move something. There are many more examples.

The symbol for torque is usually the Greek letter **tau**, **τ** .

X
#### The Greek alphabet

alpha | Α | α |

beta | Β | β |

gamma | Γ | γ |

delta | Δ | δ |

epsilon | Ε | ε |

zeta | Ζ | ζ |

eta | Η | η |

theta | Θ | θ |

iota | Ι | ι |

kappa | Κ | κ |

lambda | Λ | λ |

mu | Μ | μ |

nu | Ν | ν |

xi | Ξ | ξ |

omicron | Ο | ο |

pi | Π | π |

rho | Ρ | ρ |

sigma | Σ | σ |

tau | Τ | τ |

upsilon | Υ | υ |

phi | Φ | φ |

chi | Χ | χ |

psi | Ψ | ψ |

omega | Ω | ω |

A good, and probably familiar example of applying a torque force is using a wrench to turn a bolt. The wrench can help us to amplify the twisting force on the bolt.

You might have some experience tightening bolts. You know that they can often be turned just so tight by hand, without a wrench, but soon the wrench is needed. When using a wrench, we also know that we can apply more twisting force by pushing on the far end of the wrench as opposed to the end nearest the bolt.

Here is the **coordinate system** we use when we work with torque. The distance between the **applied force**, **F**, and the pivot point is **r**. The applied force may be applied perpendicular to the wrench or lever arm, or at some angle. If it is not perpendicular, then we need to find the perpendicular vector component of that force on the wrench, **F _{⊥} = F·cos(θ)**.

The most obvious way to change torque is to change the magnitude of the applied force. All other variables (**r**, **θ**) being constant, if **F _{2} > F_{1}**, then

If you want more torque, push harder on the wrench.

Another way to change the torque is to move the applied force closer to or further from the pivot point.

A wrench is simply a class-2 **lever**, so the closer we are to the pivot or **fulcrum**, the less the resulting twisting force – torque. In the figure, if **r _{1} > r_{2}**, then

Sometimes, when a bolt is particularly stuck, a mechanic might add a "cheater bar," a piece of pipe that slips over the wrench handle to extend it, and thus gain more torque as **r** is increased by the length of the pipe minus the overlap. Be careful, though. That's a good way to break a bolt or a wrench!

The only component of the applied force that is relevant to the torque is that part that is perpendicular to the lever (wrench in this case), **F _{⊥}**.

The two vector components of **F** are

$$ \begin{align} F_{\perp} = F \cdot sin(\theta) \\[5pt] F_{\parallel} = F \cdot cos(\theta) \end{align}$$

We don't care too much about **F _{∥}**, the component of the applied force

Putting all of these observations together, we can derive an expression for the torque:

Everything we need is there: The torque increases as **F** and **r** increase, and the torque is at a maximum when **θ = 90˚**.

**Torque** (**τ**) is **twisting force** or a force that causes rotation about one or more axes passing through the center of mass of a free object (or about some other axis if an object is so attached).

where **F** is the applied force, applied at an angle **θ** to the lever arm and at a distance **r** from the pivot axis.

The SI **units** of torque are **Newton·meters** (**N·m**). In the U.S. torque is often reported in foot-pounds (ft.-lbs.)

X
### SI units

SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units

length | meter | (m) |

mass | Kilogram | (Kg) |

time | second | (s) |

force | Newton | N |

energy | Joule | J |

Strictly-speaking, torque is a **vector product** or cross product of two vectors, the radius, **r**, and the applied force, **F** :

where all components are vectors. The **magnitude** or *size* of the torque vector (the 'amount' of torque) is

The cross product yields a new vector of the magnitude given above, and that is perpendicular both to vectors **r** and **F**. The torque vector, therefore, lies along the axis of rotation of the object to which the torque is being applied.

We use the** right-hand rule** to decide which of the two possible directions it points:

If the fingers of the right hand curl in the direction of the rotation, then the thumb will point along the direction of the torque vector. We do this by long-standing convention – we have to pick one.

The seat post bolt on a bicycle requires a torque of 35 N·m in order to remain tight. If the mechanic uses a wrench that is 30 cm long, how much force must she apply to it in order to achieve the proper torque?

**Solution**

$$\tau = F\cdot r \, sin(\theta)$$

We can rearrange that to find the force (always rearrange first before plugging in numbers!):

$$F = \frac{\tau}{r \, sin(\theta)}$$

If we assume that the force on the torque wrench will be applied at a right angle to it, then **sin(90˚) = 1**, so we can ignore the **sin(θ)** term.

Plugging in the torque and radius, we get:

$$ \require{cancel} F = \frac{35 \, N\cancel{m}}{0.3 \cancel{m}} = 117 \; N$$

Now it's worth considering a couple of things. First, notice from the torque equation that if we were to double the length of the wrench handle, we'd only need half of that 117 N of force. Second, applying the force at any angle other than 90˚ would increase the applied force needed, as the sine of the angle diminishes on either side of 90˚.

Consider two kids sitting on opposite ends of a seesaw. Their masses are m_{1} = 40 Kg and m_{2} = 32 Kg. The 40 Kg kid sits a distance of 2.8 m from the pivot point (fulcrum) of the seesaw. How far must the 32 Kg child be from the fulcrum in order to balance his friend?

**Solution**

Both forces are perpendicular to the seesaw, so the known torque is

$$ \begin{align} \tau_1 &= F_1 \cdot r_1 \\[5pt] &= (40 \, Kg)\left( 9.8 \frac{m}{s^2} \right)(2.8 \, m) \\[5pt] &= 1098 \, N\cdot m \end{align}$$

Now the torque on the other side just has to match that number. We can even ignore the signs because we have a good picture of what's going on. So we have

$$ \begin{align} \tau_2 &= F_2 r_2 \: \color{magenta}{\longrightarrow} \: r_2 = \frac{tau_2}{F_2} \\[5pt] r_2 &= \frac{1098 \, N\cdot m}{32 \, Kg \left(9.8 \frac{m}{s^2} \right)} \\[5pt] &= {\bf 3.5 \, m} \end{align}$$

Recall that $1 \, N = 1 \frac{Kg \cdot m}{s^2},$ so our units work out the right way.

That makes sense. The lighter kid will have to be farther away from the fulcrum to achieve the same torque as the heavier kid.

Consider the beam below, attached to a pivot point (red dot), and with three forces acting on it at the distances indicated from the pivot point.

- Calculate the torque produced by each force on the beam.
- Calculate the net torque on the beam.
- If a fourth force were to be applied to the beam at a distance of 0.3 m from the pivot point, what would the magnitude of that force need to be in order to achieve torque equilibrium?

**Solution**

$$ \require{cancel} \begin{align} \tau &= F \cdot r \cancel{sin(90˚)} \\[5pt] \tau_1 &= -20 \, N (0.15 \, m) = -3 \, N\cdot m \\[5pt] \tau_2 &= 15 \, N (0.25 \, m) = 3.75 \, N \cdot m \\[5pt] \tau_3 &= -5 \, N (0.35 \, m) = -1.75 \, N \cdot m \end{align}$$

The net torque is just the sum of these torques:

$$ \begin{align} \tau_1 + \tau_2 + \tau_3 &= -3 + 3.75 - 1.75 \\[5pt] &= -1 \, N\cdot m \end{align}$$

So the net torque is a downward (clockwise) twisting force of 1 N·m. To balance that, we'll need a 1 N·m torque from an unknown force applied at a distance of 3.0 m. That force is

$$ \begin{align} F &= \frac{\tau}{r} \\[5pt] &= \frac{-1.0 \, N\cdot m}{3.0 \, m} \\[5pt] &= -0.33 \, N \end{align}$$

Notice that the choice of which direction was negative and which was positive was arbitrary. Once we choose, we just have to be consistent throughout the calculations.

1. |
Two masses, m - If m
_{1}= 5 Kg, what must the mass m_{2}be in order for the mobile to balance? - Calculate the ratio of $\frac{m_1}{m_2}$ that keeps the two masses hanging horizontally.
- Suppose that m
_{1}= 11 Kg and m_{2}= 2 Kg, and the masses are arranged as shown. If a third mass, m_{3}= 5 Kg, is to be placed somewhere to balance the mobile, should it be placed to the left or to the right of the suspension point? Explain. - Calculate the exact location where m
_{3}should be placed.
## Solution |

2. |
Calculate the torque in the blue rod given the direction of the applied 50N force. ## SolutionsThe torque is $\tau = F \cdot r \, sin(\theta),$ where θ is the angle between the rotating object and the applied force. (a) The torque is $$ \begin{align} \tau_a &= F \cdot r \, sin(\theta) \\[5pt] &= 50 \, N (1.2 \, m) \cancel{sin(90˚)} \\[5pt] &= 60 \, Nm \end{align}$$ Notice that $sin(90˚) = 1,$ so when the applied force is at a right angle to the target, we can ignore the $sin(\theta)$ term. (b) The torque is $$ \begin{align} \tau_b &= F \cdot r \, sin(\theta) \\[5pt] &= 50 \, N (1.2 \, m) sin(70˚) \\[5pt] &= 56.4 \, Nm \end{align}$$ Changing the angle of the force by 20˚ from perpendicular reduces the torque. (c) The torque is $$ \begin{align} \tau_c &= F \cdot r \, sin(\theta) \\[5pt] &= 50 \, N (1.2 \, m) sin(110˚) \\[5pt] &= 56.4 \, Nm \end{align}$$ Notice that the torques in (b) and (c) are the same. Applying the 50N force at 70˚ and 110˚ both result in a force that is 20˚ off of perpendicular (θ = 90˚), just on different sides of it. The result is the same torque. |

3. |
A wood utility pole is stabilized in a vertical orientation by two wires, one attached at a height of 7 m and making a 45˚ angle with the pole, and the other attached 2 m lower on the opposite side, making a 30˚ angle to the pole. The pole is in static (non-moving) equilibrium. Calculate forc ## SolutionThe torque on the left side is $$ \begin{align} \tau &= F \, r \, sin(\theta) \\[5pt] &= 200 \, N (7 \, m) sin(45˚) \\[5pt] &= 989.9 \, N m \end{align}$$ Now if the system is in equilibrium, then the torque on the right must be the same as on the left, so the force (tension) in the wire is $$ \begin{align} F &= \frac{\tau}{r \, sin(\theta)} \\[5pt] &= \frac{989.9 \, N \cancel{m}}{5 \cancel{m} \, sin(30˚)} \\[5pt] &= 396 \, N \end{align}$$ More force is required in the shorter wire because the distance to the pivot point is shorter, and because the angle of attachment is larger than 45˚ |

4. |
If the torque required to loosen a nut on the wheel of a car has a magnitude of 42.0 Nm, what is the minimum amount of force that must be exerted at the end of a 0.35 m wrench in order to loosen the nut? ## SolutionThe minimum force will be one that is at a 90˚ angle to the wrench. Any other angle will require more force. To calculate the force, rearrange $\tau = F \cdot r$ to $$F = \frac{\tau}{r}$$ $$ \begin{align} F &= \frac{42.0 \, N\cancel{m}}{0.35 \cancel{m}} \\[5pt] &= {\bf 14.7 \, N} \end{align}$$ Notice that if we kept the $sin(\theta)$ term, we'd get $$ \begin{align} \tau &= F \, r \, sin(\theta) \\[5pt] F &= \frac{\tau}{r \cdot sin(\theta)} \end{align}$$ As θ diverges from 90˚, the value of sin(θ) decreases either way you go, so as the denominator of this equation decreases, the force increases. |

5. |
A 60 Kg window washer is standing 1.2 m from the end of a 5 m long platform that is suspended from the roof of the building by two ropes, one on each end. The platform weighs 150 N. Calculate the tension in each rope when the worker stands at this position (hint: in this context "tension" = force). ## SolutionHere is a picture of what's going on in this problem. If the person (F $$ \begin{align} com &= \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2} \\[5pt] &= \frac{\left( \frac{150}{9.81} \right) (2.5) + 60(3.8)}{15.3 + 60} \\[5pt] &= 3.54 \, m \end{align}$$ That's the distance to the center of mass from the left side. The corrected distances from the center of mass (the pivot point) look like this: The ropes suspend the entire weight of the platform plus the window washer. That's $$W = 150 \, N + 60(9.81) \, N = 738.6 \, N$$ Now the sum of all torques on the system must be zero because it's in equilibrium, so that gives us: $$ \begin{align} F_1(3.54) + 588.6(0.26)& \\[5pt] - (738.6 - F_1)(1.46) &= 0 \\[5pt] 3.54 F_1 + 1.46 F_1 + 153 - 1078 &= 0 \\[5pt] 5.0 F_1 &= 925 \\[5pt] F_1 &= 185 \, N \end{align}$$ Then it follows that $F_2 = 554 \, N.$ |

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