drawing pictures 2.1.2a free body diagrams and net force
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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