Drawing Pictures

2.1.2A Free Body Diagrams and Net Force

Free Body Diagrams• VECTOR diagrams

• Show ALL FORCES acting on an object

• Must be properly labeled

Fg

FN

Equilibrium

Determining Net Force• equilibrium net force = 0 acceleration = 0

– MOTIONLESS or CONSTANT VELOCITY

• net force ≠ 0– SPEEDING UP or SLOWING DOWN

Determining Net Force – Top View• Consider each dimension separately then

combine using Pythagorean Theorem.

F1 = +7.0 N

X Y

+7.0 N

0 N

Net force = +7.0 Na = 3.5 m/s2

+7.0 N

2 kg object

Not in notes

Determining Net Force – Top View• Consider each dimension separately then

combine using Pythagorean Theorem.

X Y

+7.0 N-3.0 N

0 N+4.0 NF2 = -3.0 N F1 = +7.0 N

Net force = +4.0 Na = 2.0 m/s2

2 kg object

Not in notes

Determining Net Force – Top View• Consider each dimension separately then

combine using Pythagorean Theorem.

X Y

+7.0 N-3.0 N

+3.0 N+4.0 NF2 = -3.0 N F1 = +7.0 N

F3 = +3.0 N +3.0 N

Net force = 5.0 Na = 2.5 m/s2

2 kg object

Not in notes

Determining Net Force – Top View• Consider each dimension separately then

combine using Pythagorean Theorem.

X Y

+7.0 N-3.0 N

-2.0 N+4.0 NF2 = -3.0 N F1 = +7.0 N

F3 = +3.0 N +3.0 N-5.0 N

F4 = -5.0 NNet force = 4.5 N

a = 2.25 m/s2

2 kg object

Notes

• A 30 newton force and a 20 newton force act concurrently on an object.– What are the minimum/maximum net force can these two

forces produce? At what angle between the forces does each happen?

– What net force is produced when the angle between the two forces is 90°?

Example #1

min = 10 N occurs at 180°max = 50 N occurs at 0°

36 N

Determining Net Force – Side View• Consider each dimension separately.• On a flat surface total vertical force = 0.

X Y

+8.0 N

0 N+8.0 NF = +8.0 N

-Fg

+FN

frictionless Net force = +8.0 Na = 4.0 m/s2

2 kg object

Not in notes

Determining Net Force – Side View• Consider each dimension separately.• On a flat surface total vertical force = 0.

Hor. Vert.

+8.0 N-2.0 N

0 N+6.0 NF = +8.0 N

-Fg

+FN

Net force = +6.0 Na = 3.0 m/s2

2 kg object

Ff = -2.0 N

Notes

• A 50 kilogram object is pushed along a flat, frictionless surface with a constant force of 100 newtons.– Sketch this object. Include all forces with labels and

quantities.

– What rate of acceleration will the crate experience?

Example #2

Determining Net Force – Vertical• Object moving straight up or down total

horizontal force = 0.

Hor. Vert.

-20 N

-20 N0 N

Net force = -20 Na = -9.81 m/s2

20 N object in freefall

Fg = -20 N

Not in notes

• What is the net force acting on a 3000 newton rocket if its engine produces an upward thrust of 4500 newtons?

Example #3

Fg

Fthrust

1500 N upward

End of 2.1.2A - PRACTICE