Science Starter!Draw a free-body diagram for:

1) A chair at rest on the floor.

2) A ball rolling to the right and slowing down across a grassy field.

1.)

2.)

according to Newton’s 1st Law:

- objects at rest remain at rest unless a force is applied to move them

- objects in motion stay in motion unless a force is applied to change their speed or direction

Remember that….

Newton’s Second Law

The net force applied to an object to change its state of motion is directly

proportional to the object’s mass and resulting acceleration of the object.

Newton’s 2nd Law (formula form)

FNet = ma

Fnet (or ΣF): “Net Force” - The vector sum of all forces acting on an object [Newtons (N)]

m: mass (kg)

a: acceleration (m/s2)

Weight vs. MassWeight: “Fg” force of gravity pulling on an

object

F = ma a acceleration“g” acceleration due to gravity (9.8 m/s2)

THEREFORE:

Fg = mg

Example

A ball has a mass of 8.0 kg. What is the ball’s weight on earth ?

(1) m = 8.0 kg g = 9.8 m/s2

Fg = ?

(2) Fg = mg

(3) Fg = (8.0 kg) (9.8 m/s2)

(4-5) Fg = 78.4 N

Net ForceForces in “x” direction can be added,Forces in “y” direction can be added:

Can be written as: FNETx

OR Σ Fx

FNETy OR Σ Fy

CHEAT: Direction of acceleration is the POSITIVE DIRECTION

Example 1A 10 kg wagon is being pulled to the right with 20 N of force. Between the tires and

the ground, there is 12 N of friction.

a) Draw a free-body diagram.b) Write an “FNET Equation” for the vertical and horizontal directions.c) Determine the acceleration of the wagon.

Σ Fy = FN – Fg Σ Fx = F – Ff

(0) = FN – (10)(9.8) m(a) = F – Ff

10 (a) = (20) – (12)

10 (a) = 8a = 0.8 m/s2

Two students are playing tug-of-war over a 15 kg crate. Joe pulls to the right with 40 N of force, while Bob pulls to the left with 60 N of force. The frictional force between the crate and the

ground is 5 N.

a) Draw a free-body diagram.b) Write an “FNET Equation” for the vertical and horizontal directions.c) Determine the acceleration of the crate.

Example 2

Σ Fy = FN – Fg Σ Fx = FBob – Ff – FJoe

(0) = FN – (15)(9.8) m(a) = FBob – Ff – FJoe

15 (a) = (60) –

(5) – (40)

15 (a) = 15

a = 1.0 m/s2

Hooke’s Law: the Spring Force

Robert Hooke:

Fs = kx

Fs: Spring Force (N)

k: Spring Constant (N/m)x: Stretch distance (m)

Example 3A 5 kg mass hangs from a spring of spring constant 120 N/m. How far will the spring

stretch once the mass is at rest?

a) Draw a free-body diagram of the mass.b) Write an “FNET Equation” for the vertical and horizontal directions.c) Determine the stretch distance of the spring. How many centimeters is this?

Σ Fy = Fs – Fg Σ Fx = 0

(m)(a) = (k)(x) – (m)(g)5 (0) = (120)(x) – (5)(9.8)0 = 120 (x) – 4949 = 120 xx = 0.408 m = 40.8 cm