Newton's Laws of Motion

Issac Newton(1643-1727) made many profound contributions to science. He was an important figure in the Scientific Revolution. He worked on optics. He discovered universal gravitation. He also shares credit for inventing calculus.

In his work Philosophiæ Naturalis Principia Mathematica, Newton formulated his three laws of motion. These laws make up the foundation of classical mechanics. They predict how forces cause acceleration.

Newton's First Law: Inertia

"When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by an external force."

Objects keep moving in a straight line at a constant velocity unless they have a force acting on them.

Objects with mass have inertia. Inertia is the resistance of any physical object to any change in velocity. The more mass an object has the more force is required to change its motion.

Objects in space keep moving without slowing down. Why do things on Earth slow down so quickly?

Newton's Second Law: Force

"The vector sum of the external forces F on an object is equal to the mass m of that object multiplied by the acceleration vector of the object."

$$\sum F=ma$$ $$\sum F=F_{1}+F_{2}+F_{3}+...$$

$$\sum$$ = The greek letter Sigma represents a summation of numbers. It means add up all the forces to get a net force.

$$F$$ = force [N, Newtons, kg m/s²] vector
a push or a pull

$$m$$ = mass [kg, kilogram]
the measure of an object's resistance to acceleration

$$a$$ = acceleration [m/s²] vector
the change in velocity every second

Example: A soccer ball has a mass of 0.43 kg. Find the acceleration of the ball when it experiences a net force (total force) of 1 N from air friction.
solution
• $$\text{no units}$$ $$F=ma$$ $$\frac{F}{m}=a$$ $$\frac{1}{0.43}=a$$ $$2.33 \, \mathrm{\tfrac{m}{s^{2}}}=a$$
• $$\text{with units}$$ $$F=ma$$ $$\frac{F}{m}=a$$ $$\frac{1 \, \mathrm{N}}{0.43 \, \mathrm{kg}}=a$$ $$\frac{1 \, \mathrm{kg \frac{m}{s^2}}} {0.43 \, \mathrm{kg} }=a$$ $$2.33 \, \mathrm{\tfrac{m}{s^2}}=a$$
Example: A 100 kg boat at rest is being pushed left with a 2000 N force from the wind. The water current is producing a force of 1900 N to the right. How will the boat accelerate? How far will the boat go in 6 s?
solution $$\text{Vectors pointed down or left are negative.}$$ $$\sum F=ma$$ $$-2000+1900=(100)(a)$$ $$-100=(100)(a)$$ $$-1 \mathrm{\tfrac{m}{s^{2}}}= a$$ $$\text{The boat will accelerate to the left.}$$ $$u = 0$$ $$\Delta x = ?$$ $$\Delta t = 6\mathrm{s}$$ $$\Delta x = u\Delta t+ \tfrac{1}{2} a \Delta t^{2}$$ $$\Delta x = (0)(6)+\tfrac{1}{2} (-1)(6)^{2}$$ $$\Delta x = \tfrac{1}{2} (-1)(6)^{2}$$ $$\Delta x = -18 \, \mathrm{m}$$ $$\text{The boat will move 18 meters to the left.}$$
Example: If I apply a force to a mass it accelerates. What happens if I apply a larger force to the same mass? Use proportions.
solution $$\text{m is constant}$$ $$\text{F and a are directly proportional}$$ $${\Uparrow \atop F} {\atop =} { \atop m} {\Uparrow \atop a}$$ $$\text{Increasing F will increase a}$$
Example: If I apply a 100 N force to a mass it accelerates at 2m/s². What happens if I double the mass while keeping the force the same? Use proportions.
solution $$\text{F is constant}$$ $$\text{m and a are inversely proportional}$$ $${ \atop F} {\atop =} { \Downarrow \atop m} {\Uparrow \atop a}$$ $$\text{doubling m will half a}$$ $$a = 1 \, \mathrm{\tfrac{m}{s^2}}$$
Example: Use the free body diagram to calculate the acceleration for both the horizontal and the vertical.
solution $$\text{vertical}$$ $$\sum F=ma$$ $$22 {\color{#aaa} \, \mathrm{N}} - 85{\color{#aaa} \, \mathrm{N}} = (16 {\color{#aaa} \, \mathrm{kg}})a$$ $$-63 {\color{#aaa} \, \mathrm{N}} = (16 {\color{#aaa} \, \mathrm{kg}})a$$ $$-3.9 {\color{#aaa} \, \mathrm{\tfrac{m}{s^2}}} = a$$
$$\text{horizontal}$$ $$\sum F=ma$$ $$78{\color{#aaa} \, \mathrm{N}}+55{\color{#aaa} \, \mathrm{N}}-140{\color{#aaa} \, \mathrm{N}} = (16 {\color{#aaa} \, \mathrm{kg}})a$$ $$-7{\color{#aaa} \, \mathrm{N}} = (16 {\color{#aaa} \, \mathrm{kg}})a$$ $$-0.43 {\color{#aaa} \, \mathrm{\tfrac{m}{s^2}}} = a$$

Newton's Third Law: Equal and Opposite Force Pairs

"When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body."

If you push on something it will push back with the same force, but in the opposite direction. Forces always come in equal, but opposite pairs.

Example: You push down on the ground with a 1000 N force. Describe the magnitude and direction of the force the ground pushes on you.
solution

The force the ground applies is equal and opposite to the force you apply on the ground. The magnitude is 1000 N. The direction is up.

Example: A boxer's glove applies a 140 N force to the face of another boxer. Describe the magnitude of the force the face applies to the glove.
solution

The force is equal and opposite. The magnitude is 140 N.

Example: You (100 kg) are floating away from your spaceship. In order to return to the ship you throw your 2 kg space laser away. As you are throwing the space laser, it accelerates at 40 m/s². How much acceleration do you feel during the throw?
solution

The force the space laser feels is equal and opposite to the force you feel.

$$\text{space laser}$$ $$F = ma$$ $$F=(2)(40)$$ $$F=80 \, \mathrm{N}$$
$$\text{you}$$ $$F = ma$$ $$-80 = 100a$$ $$-0.8 \mathrm{\tfrac{m}{s^2}}=a$$
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.

$$F_{\mathrm{squirrel}}=5.2 \, \mathrm{N} \text{ left}$$