# Motion - Dynamics

**Dynamics** is the study of *why* things move, in contrast to kinematics, which is concerned with describing the motion of objects.
An object's motion typically is described using Newton's Laws of Motion.

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Topic: High School Physics |

## Newton's Laws of MotionEdit

### Newton's 1st Law of MotionEdit

Newton's First Law is often stated: "An object at rest will tend to stay at rest, or an object in motion will tend to stay in uniform motion in a straight line unless acted on by an external unbalanced force." This means that an object must have an unbalanced force acting on it for it's motion to change.

#### Example 1Edit

Consider a bus moving steadily at 100 km/h along a highway. Although we can assume that many forces are acting on the bus, such as weight (force due to gravity), air resistance and/or friction with the highway, we know via Newton's 1st Law of Motion that the sum of the forces acting on the bus is zero, as it is at a constant velocity.

#### Example 2Edit

Consider a truck going 40 km/hr down a straight road (it is important that the road is straight in this example!). If a grocery bag is sitting in the passenger seat and the driver slams on the brakes, the groceries will typically fly off of the seat, into the glove compartment, or onto the floor. This is because the brakes are creating an unbalanced force slowing the truck, but not the grocery bag. But why doesn't the driver also get thrown out of their seat? The driver is (hopefully) wearing a seat belt, which is connected to the truck. The tension of the seat belt is providing a similar force to the driver.

#### Applications to Solving Physics ProblemsEdit

In any situation which you have an object at rest or at constant velocity, you can take the sum of all forces acting on it to be zero, which is to say the net force acting on the object is zero.

#### See Newton's First Law in ActionEdit

A good experiment to illustrate the Newton's First Law of Motion is to put a book on a skateboard. Push or run behind the skateboard/book to accelerate them up to a good speed, then suddenly stop the skateboard, but do not touch the book. Assuming the book is not incredibly heavy, it should continue moving forward after the skateboard stops. This phenomenon can be explained using Newton's First Law.

### Newton's 2nd Law of MotionEdit

Newton's 2nd Law of Motion states:

"the rate of change of the momentum of an object is directly proportional to the resultant force acting upon it".

Written mathematically, this gives

this can easily be reduced to F = ma by using differentials

in a constant mass system

and so

as

Therefore, in a system of constant mass, the acceleration of an object is directly proportional to the resultant force acting upon it.

This formula is an experimental result. You can find it for yourself if you have some means of measuring acceleration, such as a tickertape timer or sonar attachment for a calculator to measure velocity continuously.

In the experiment, various forces are applied to a wheeled cart or glider on a frictionless air track. The easiest way to get a known force is to use the force of gravity on a hanging weight with a pulley to change the force from horizontal to vertical. Each 100 grams of mass has a force of gravity of about 1 Newton.

The tickertape is pulled through a timer that marks a dot on it every tenth of a second. From the distance between dots, the velocity can be calculated. From the change in velocity from one pair of dots to the next, the acceleration can be calculated. The experiment is repeated with various hanging masses causing different pulling forces and the acceleration is measured from the recorded motion. Graphing the experimental values for the applied force versus the resulting accelerations produces a straight line graph to within experimental accuracy. The slope is equal to the mass of the moving system to within experimental error. The formula for the graph is therefore F = ma. Force has units of Newtons where a Newton (N) is equal to a kg·m/s².

#### Example 1Edit

A 1000 kg car accelerates at 0.5 m/s². Calculate the force that must be applied to it.

- Solution

#### Example 2Edit

A 0.085 kg bullet is fired from a rifle and emerges with a speed of 400 m/s. Assuming that the bullet has constant acceleration over the 0.5 m length of gun barrel, calculate the force on the bullet.

**Solution:**
This is one of those tricky problems when the time is not known so we can't use a = Δv/Δt. We could sketch a v vs t graph where we know the area beneath is the distance 0.5 m:

Knowing the time for the bullet to accelerate in the rifle, we can find the acceleration and then the force:

### Newton's 3rd Law of MotionEdit

Newton's Third Law of Motion is often stated as "For every action there is an equal and opposite reaction." This means that for every force exerted, the exerter has an force of same magnitude but opposite direction from the exertee. More plainly put, if object A applies a force on object B, then B will apply an equal force on A in the opposite direction.

#### Example 1Edit

Consider a car traveling. If the car accelerates due to a 500 N force on it there is another object that experiences a 500 N force in the opposite direction. In this case, it would be the road. Similarly, an aircraft accelerates because it applies a backward force on the air (via a propeller or turbine), and the air applies a forward force on the aircraft. In space, where there is no air, a rocket are typically used - it applies a backwards force on its exhaust material and the exhaust material applies a forward force on the rocket.

Often there are multiple forces involved. When a person shoots a rifle, a force is obviously applied to the bullet. It is the hot gunpowder gases that push on the bullet, and the bullet pushes back on the gases. The gases push back on the gun and the gun pushes forward on the gases. The gun pushes back on the person holding it, and the person pushes forward on the gun. You could continue this to the person pushing back on the Earth and the Earth pushing forward on the gun.

#### Applications to Solving Physics ProblemsEdit

##### Action/Reaction PairsEdit

Consider the Earth and a ball. The force from the Earth on the ball, and the force from the ball on the Earth are of equal magnitude, but opposite directions.

The Earth and a ball in the air exert equal and opposite forces on each other. The Earth and ball in this situation are said to be "Action/Reaction" pairs.

However, consider a 100 kg mass sitting at rest on a table. There is a force downwards from the mass on the table (weight) and a force upwards from the table on the mass (normal force). However, weight and normal force are not Action/Reaction Pairs, as the weight is not created by the normal force. The true normal force in this example are the mass and the Earth, as the Earth is applying a gravitational force on the mass (weight) and the mass is applying the same magnitude force in a opposite direction.

Action/Reaction Pairs are handy to solving many Physics Exercises.