# Forces

Free body diagrams are a technique that helps visualize the force vectors used to solve Newton's second law.

A free body diagram shows all the force vectors on an object. Mass is typically written inside a box, with force vectors pointed away from the box.

Example: Draw a free body diagram for a 83 kg person holding onto a rope in a game of "Tug of war". The rope is pulling them with a force of 520 N left. The person's feet are countering that with 540 N to the right.
solution
Example: Use the free body diagram to calculate the mass's acceleration.
solution $$\sum F=ma$$ $$-200\,\mathrm{N}+300\,\mathrm{N}+500\,\mathrm{N}=ma$$ $$600\,\mathrm{N}=(2\,\mathrm{kg})a$$ $$300 \mathrm{\tfrac{m}{s^2}}=a$$
Example: The book The Martian by Andy Weir (mass = 0.09 kg) is resting on a table (acceleration = zero). The force of gravity produces a downwards force of 0.34 N. What force is required to keep the book at rest on the table?
solution $$\sum F=ma$$ $$F_{N} + F_{g} = ma$$ $$F_{N} - 0.34\, \mathrm{N} = (0.09\, \mathrm{kg})(0)$$ $$F_{N} - 0.34\,\mathrm{N} = 0$$ $$F_{N} = 0.34 \, \mathrm{N}$$
Example: Use the free body diagram to calculate the acceleration for both the horizontal and the vertical.
solution $$\text{vertical}$$ $$\sum F=ma$$ $$22 \, \mathrm{N} - 85 \, \mathrm{N} = (16 \, \mathrm{kg})a$$ $$-63 \, \mathrm{N} = (16 \, \mathrm{kg})a$$ $$-3.9 \, \mathrm{\tfrac{m}{s^2}} = a$$
$$\text{horizontal}$$ $$\sum F=ma$$ $$78 \, \mathrm{N}+55 \, \mathrm{N}-140 \, \mathrm{N} = (16 \, \mathrm{kg})a$$ $$-7 \, \mathrm{N} = (16 \, \mathrm{kg})a$$ $$-0.43 \, \mathrm{\tfrac{m}{s^2}} = a$$
Example: A red crate is falling, but a parachute is slowing its acceleration to only 2 m/s² down. The force of gravity is 833 N down. The parachute provides 663 N up. Draw a free body diagram and use it to find the mass of the crate.
solution $$\sum F=ma$$ $$F_1+F_2=ma$$ $$-833+663 = m(2)$$ $$\frac{-170}{2} = \frac{m(2)}{2}$$ $$-85 \, \mathrm{kg} = m$$

A negative mass doesn't make sense. Where did we mess up?
Oh, the acceleration is negative, because it points down.

$$\frac{-170}{-2} = \frac{m(-2)}{-2}$$ $$85 \, \mathrm{kg} = m$$

# The Force of gravity

Mass is a measure of an object's inertia. Mass also determines the strength of gravity. Because of gravity all objects are attracted to each other, but we mostly notice the attraction towards the Earth because it is so large and so close.

There is a special word to describe the direction of gravity, down.

# $$F_g=mg$$

$$F_g$$ = the force of gravity, weight [N, newtons, kg m/s²] vector
$$m$$ = mass [kg]
$$g$$ = acceleration of gravity on Earth = 9.8 [m/s²] vector

The force of gravity depends only on the mass of the object because on the surface of the Earth acceleration from gravity is the same for all objects. If you aren't on the surface of the Earth, there is a different way to calculate gravity.

Question: Why do feathers fall slower than bricks?
solution

Air friction produces a force that opposes motion. Feathers have a large surface area compared to their small mass so they have more air friction.

The acceleration of 9.8 m/s² on the surface of the Earth is just an approximation. The gravity of Earth changes a bit depending on where you are.

Table: Comparative gravities in various cities around the world
Location Acceleration in m/s² Acceleration in ft/s²
Amsterdam 9.813 32.19
Athens 9.800 32.15
Auckland 9.799 32.15
Bangkok 9.783 32.1
Brussels 9.811 32.19
Buenos Aires 9.797 32.14
Calcutta 9.788 32.11
Cape Town 9.796 32.14
Chicago 9.803 32.16
Copenhagen 9.815 32.2
Frankfurt 9.810 32.19
Havana 9.788 32.11
Helsinki 9.819 32.21
Istanbul 9.808 32.18
Jakarta 9.781 32.09
Kuwait 9.793 32.13
Lisbon 9.801 32.16
London 9.812 32.19
Los Angeles 9.796 32.14
Manila 9.784 32.1
Mexico City 9.779 32.08
Montréal 9.789 32.12
New York City 9.802 32.16
Nicosia 9.797 32.14
Oslo 9.819 32.21
Ottawa 9.806 32.17
Paris 9.809 32.18
Rio de Janeiro 9.788 32.11
Rome 9.803 32.16
San Francisco 9.800 32.15
Singapore 9.781 32.09
Skopje 9.804 32.17
Stockholm 9.818 32.21
Sydney 9.797 32.14
Taipei 9.790 32.12
Tokyo 9.798 32.15
Vancouver 9.809 32.18
Washington, D.C. 9.801 32.16
Wellington 9.803 32.16
Zurich 9.807 32.18

Question: What factors might explain why the measured acceleration of gravity changes in different locations on the surface of Earth?
solution

• Gravity is slightly weaker at higher elevation because you are farther from the center of the Earth.

• Near the equator gravity feels weaker because the Earth's rotation adds a centrifugal force.
• Example: The Three-Body Problem by Cixin Liu has a mass of 0.44 kg. What force of gravity does the book have?
solution $$F_g=mg$$ $$F_g=(0.44 \, \mathrm{kg} )(9.8\, \mathrm{\tfrac{m}{s^2}})$$ $$F_g=4.3 \, \mathrm{N}$$ Example: How much mass does the book Cat's Cradle by Kurt Vonnegut have if it feels a force of gravity of 1.8 N?
solution $$F_g=mg$$ $$\frac{F_g}{g}=m$$ $$\frac{1.8 \,\mathrm{N}}{9.8\, \mathrm{\tfrac{m}{s^2}}}=m$$ $$0.18 \, \mathrm{kg}=m$$

### Weight and Mass

Another word for the force of gravity is weight. An object on the Moon would weigh less than it does on Earth because of the lower gravity, but it would still have the same mass.

$$F_g = \mathrm{weight}$$

Earth's gravity does extend into space, but it decreases with distance. It is about ~90% for astronauts in orbit around the Earth, but they don't notice any gravity because they are in a freefall.

Freefall means that you are just letting gravity accelerate you without any opposing forces. To keep from falling we are careful to always counter the force of gravity. This can be done with a parachute, or a jet pack, or just the ground. Question: The Name of the Wind by Patrick Rothfuss has a mass of 0.34 kg. How does its mass and weight change in a freefall on Earth?

An object's mass doesn't change when it is falling.

Weight just means the force of gravity, which also doesn't change in a short freefall. Example: The hardcover version of Seveneves by Neal Stephenson has a mass of 0.95 kg. What is the force of gravity felt by the book on Earth? What about on the Moon?
Local Massive Objects Surface Gravity
name g (m/s²)
Sun 275
Mercury 3.7
Venus 8.9
Earth 9.8
Moon 1.6
Mars 3.7
Jupiter 25.8
Saturn 10.4
Uranus 8.7
Neptune 11.2
solution

Weight and force of gravity mean the same thing. A planet's gravity field determines your weight, but not your mass.

• $$\text{Earth}$$ $$F_{g}=mg$$ $$F_{g}=(0.95)(9.8)$$ $$F_{g}=9.31 \, \mathrm{N}$$
• $$\text{Moon}$$ $$F_{g}=mg$$ $$F_{g}=(0.95)(1.6)$$ $$F_{g}=1.52 \, \mathrm{N}$$
Converting kilograms (kg) into pounds (lbs) $$1 \, \mathrm{kg} = 2.2\, \mathrm{lbs} \quad \scriptsize \text{(On Earth)}$$ $$1 \, \mathrm{N} = 0.2248\, \mathrm{lbs}$$

Pounds are a unit of force and and kilograms are a unit of mass. You can't convert directly between them because they are different concepts, but you can use the force of gravity equation to find a conversion that works for only Earth's surface. Example: Uprooted by Naomi Novik is resting on a table. The shipping weight is 1.2 pounds. What is the book's mass in kilograms on Earth? On Mars?
solution

Mass doesn't depend on gravity, so it's the same everywhere.

$$1.2\,\mathrm{lbs} \left( \frac{1\,\mathrm{kg}}{2.2\,\mathrm{lbs}} \right)= 0.\overline{54}\,\mathrm{kg}$$ $$m = 0. \overline{54} \, \mathrm{kg}$$

What is the weight of the book in Newtons on Earth? On Mars?
solution
• $$\text{weight on Earth}$$ $$F_{g}=mg$$ $$F_{g}=(0.\overline{54})(9.8)$$ $$F_{g}=5.35\, \mathrm{N}$$
• $$\text{weight on Mars}$$ $$F_{g}=mg$$ $$F_{g}=(0.\overline{54})(3.711)$$ $$F_{g}=2.02\, \mathrm{N}$$

### The Normal Force

Typically a normal force will balance the force of gravity to keep an object from accelerating up or down.

A normal force occurs when two objects are in contact. It is perpendicular to the point of contact. A normal force prevents objects from passing through each other.

A normal force will scale to a value that will keep the net force and acceleration zero.

Normal forces come from the combined effect of electromagnetic forces and the Pauli exclusion principle.

The electromagnetic force allows chemical bonds to form. These bonds give solid matter its rigid structure, which is required for normal forces.

At the atomic scale, particles can't pass through each other primarily because of a quantum mechanical effect called the Pauli exclusion principle . The Pauli exclusion principle is mostly responsible for keeping particles, like electrons, separate.

Press E to activate the mass. Then press WASD to apply forces to the mass. Imagine that gravity is pointed towards the bottom of the page.

Question: What force keeps the mass from exiting the screen?

The normal force.

Question: Why does pressing A and D at the same time do nothing?

The left and right force cancel each other out.

Example: You are accelerating up in an elevator at 2 m/s². If your mass is 100 kg, what is the normal force you feel from the elevator?
solution $$F_{g}=mg$$ $$F_{g}=(100)(-9.8)$$ $$F_{g}=-980\, \mathrm{N}$$
$$\sum F=ma$$ $$F_{N}-F_{g}=ma$$ $$F_{N} - 980=(100)(2)$$ $$F_{N}=1180\, \mathrm{N}$$
Example: A 20 kg box is at rest on a horizontal sidewalk. Find the force of gravity and the normal force on the box.
solution

In the simple case of a flat horizontal surface with no vertical acceleration the force of gravity will always be equal and opposite to the normal force.

$$F_{g}=mg$$ $$F_{g}=(20)(9.8)$$ $$F_{g}=196\, \mathrm{N}$$
$$\sum F=ma$$ $$F_{N}-F_{g}=ma$$ $$F_{N} - 196=(20)(0)$$ $$F_{N}=196 \, \mathrm{N}$$
Example: A 20 kg box is at rest on a steep sidewalk. The sidewalk is at an angle 20 degrees from horizontal. Find the force of gravity and the normal force on the box. What is the acceleration of the box? (assume no friction)
solution
• $$F_{g}=mg$$ $$F_{g}=(20)(9.8)$$ $$F_{g}=196 \, \mathrm{N}$$

Separate the gravity vector into components parallel and perpendicular to the ground. Acceleration is zero in the perpendicular direction.

• $$\text{perpendicular to ground}$$

$$F_{g\perp}=F_{g}\cos(20)$$ $$F_{g\perp}=(196)\cos(20)$$ $$F_{g\perp}=184 \, \mathrm{N}$$
$$\sum F_{\perp}=ma$$ $$-F_{g\perp}+F_{N}=ma$$ $$F_{N}=ma+F_{g\perp}$$ $$F_{N}=(20)(0) + 184$$ $$F_{N}=184 \, \mathrm{N}$$
• $$\text{parallel to ground}$$

$$F_{g\parallel}=F_{g}\sin(20)$$ $$F_{g\parallel}=(196)\sin(20)$$ $$F_{g\parallel}=67 \, \mathrm{N}$$
$$\sum F_{\parallel}=ma$$ $$F_{g\parallel}=ma$$ $$(67)=(20)a$$ $$3.35 \, \mathrm{\tfrac{m}{s^{2}}}=a$$

### Tension Forces

When a person is walking a dog they are able to apply a force on the dog with a leash. They use the tension on the leash to transfer that force from their hand to the dog.

Tension is the pulling force from a chain, string, or rope. Tension is useful for transferring a force over a distance. In most situations, the tension is the same for both ends.

Example: A helium balloon is attached to a 0.5 g paper clip. If the balloon and paper clip are accelerating up at 0.023 m/s², what is the tension on the paper clip from the balloon in Newtons?
solution $$\sum F=ma$$ $$-F_{g} + T = ma$$ $$T = ma + F_{g}$$ $$T = ma + mg$$ $$T = (0.0005\, \mathrm{kg})(0.023\, \mathrm{\tfrac{m}{s^2}})+ (0.0005\, \mathrm{kg}) (9.8\, \mathrm{\tfrac{m}{s^2}})$$ $$T = 0.0049115\, \mathrm{N}$$
Example: A 100 kg person is pulling a 10 kg crate with a rope (ignore the mass of the rope). Both the person and the crate are accelerating to the left at 0.1 m/s². What force is the person producing in order to accelerate to the left?
solution

The tension force is equal and opposite for the person and crate.

$$\text{crate on right}$$ $$\sum F=ma$$ $$T = 10 (-0.1)$$ $$T = \color{#f05}-1 \, \mathrm{N}$$
$$\text{person on left}$$ $$\sum F=ma$$ $$F_{\mathrm{person}} + T = ma$$ $$F_{\mathrm{person}} {\color{#f05}+ 1 \, \mathrm{N}} = (100 \, \mathrm{kg})(-0.1\, \mathrm{\tfrac{m}{s^2}})$$ $$F_{\mathrm{person}} + 1\, \mathrm{N} = -10\, \mathrm{N}$$ $$F_{\mathrm{person}} = -11\,\mathrm{N}$$
Example: You are walking your dog with the leash at a 45 degree angle down towards the dog. Neither your nor the dog are accelerating. The dog is pulling on the leash forward with a force of 100 N. Calculate the x part of the tension force on the leash. Then calculate the total tension force.
solution $$\text{dog: horizontal}$$ $$\sum F=ma$$ $$-T_x + F_{\mathrm{dog}} = 0$$ $$-T_x + 100 \, \mathrm{N} = 0$$ $$T_x= 100 \,\mathrm{N}$$

The 100 N is only the x-part of the tension force vector. We can find the total tension force with the Pythagorean theorem. At 45° the x and y parts of the force are the same.

$$T^2 = 100^2+100^2$$ $$T^2 = 20\,000$$ $$T = 141 \, \mathrm{N}$$
Example: A 0.4 kg squirrel is pulling a 10 kg box with a string. The squirrel and box are accelerating to the left at 0.5 m/s². What is the force of tension on the string? What force is the squirrel producing in order to accelerate to the left?
solution $$\text{box}$$ $$\sum F=ma$$ $$F_{\mathrm{tension}}=(10)(-0.5)$$ $$F_{\mathrm{tension}}=-5\, \mathrm{N}$$

The tension force on the string is equal but opposite for the squirrel and box.

$$\text{squirrel}$$ $$\sum F=ma$$ $$F_{\mathrm{squirrel}} + F_{\mathrm{tension}}=ma$$ $$F_{\mathrm{squirrel}} + 5=(0.4)(-0.5)$$ $$F_{\mathrm{squirrel}}=-0.2-5$$ $$F_{\mathrm{squirrel}}=-5.2 \, \mathrm{N}$$

The negative sign tells us that the squirrel's force is pointed left.

$$F_{\mathrm{squirrel}}=5.2 \, \mathrm{N} \text{ left}$$
Example: Use the diagram to predict the acceleration of the masses.
Assume no friction, no air resistance, Earth gravity, and a massless rope.
hint

Both free body diagrams share the same tension and acceleration. Build two equations with Newton's second law and solve a system of equations for T and a. While T and a are the same magnitude they have different directions, so watch the sign of acceleration in particular.

This Khan academy video covers this problem in more depth.

solution $$\text{20kg box: horizontal}$$ $$\sum F=ma_x$$ $$T = ma_x$$ $$T = \color{#d3a}20a_x$$
$$\text{10kg box: vertical}$$ $$\sum F=ma_y$$ $$T - F_g = ma_y$$ $$T - mg = ma_y$$ $$T - (10)(9.8) = 10a_y$$
$${\color{#d3a}20a_x} - 98 = 10a_y$$
$$a_x = -a_y$$
$${\color{#d3a}-20a_y} - 98 = 10a_y$$ $$-98 = 10a_y + 20a_y$$ $$-98 = 30a_y$$ $$-3.2\overline{6} \, \mathrm{\tfrac{m}{s^2}} = a_y$$
solution (treating system as one mass) $$\text{gravity only pulls on the 10kg part}$$ $$F_g = m g$$ $$F_g = (10 \, \mathrm{kg})(9.8\, \mathrm{\tfrac{m}{s^2}})$$ $$F_g = \color{#d3a}98 \, \mathrm{N}$$
$$\text{use both masses to find acceleration in the vertical}$$ $$\sum F=ma$$ $$F_g = (10\, \mathrm{kg}+20\, \mathrm{kg}) a$$ $$F_g = (30\, \mathrm{kg}) a$$ $$98\, \mathrm{N} = (30\,\mathrm{kg}) a$$ $$-3.2\overline{6} \, \mathrm{\tfrac{m}{s^2}}= a$$
Example: Two blocks are attached to each other with a rope hanging over a wheel. Find the acceleration of each body. Ignore friction, and assume the wheel and rope have a negligible mass.
solution
$$\text{10 kg block}$$ $$\sum F = ma$$ $$T-mg = ma$$ $$T-(10)(9.8) = ma$$ $$T-98 = 10a$$ $$T = 10a+98$$
$$\text{25 kg block}$$ $$\sum F = ma$$ $$T-mg = ma$$ $$T-(25)(9.8) = 25a$$ $$T-245 = 25a$$ $$T = 25a+245$$

We have two variables and two equations, this means we can plug one equation into the other. The Tension is the same for each body. The accelerations are the same, but in opposite directions, so we need to make one acceleration negative.

$$T = 10({\color{#f05}-a})+98$$
$$T = 25a+245$$
$$10({\color{#f05}-a})+98 = 25a+245$$ $$-35a = 147$$ $$a = -4.2 \, \tfrac{m}{s^2}$$

The 25kg object is falling down at 4.2 m/s². The 10kg is rising up at 4.2 m/s².