2.1.2A Free Body Diagrams and Net Force

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Presentation transcript:

2.1.2A Free Body Diagrams and Net Force 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 Equilibrium FN Fg

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. X Y 2 kg object +7.0 N F1 = +7.0 N +7.0 N 0 N Net force = +7.0 N a = 3.5 m/s2 Not in notes

Determining Net Force – Top View Consider each dimension separately then combine using Pythagorean Theorem. X Y 2 kg object +7.0 N -3.0 N F2 = -3.0 N F1 = +7.0 N +4.0 N 0 N Net force = +4.0 N a = 2.0 m/s2 Not in notes

Determining Net Force – Top View Consider each dimension separately then combine using Pythagorean Theorem. X Y 2 kg object F3 = +3.0 N +7.0 N -3.0 N +3.0 N F2 = -3.0 N F1 = +7.0 N +4.0 N +3.0 N Net force = 5.0 N a = 2.5 m/s2 Not in notes

Determining Net Force – Top View Consider each dimension separately then combine using Pythagorean Theorem. X Y 2 kg object F3 = +3.0 N +7.0 N -3.0 N +3.0 N -5.0 N F2 = -3.0 N F1 = +7.0 N +4.0 N -2.0 N Net force = 4.5 N a = 2.25 m/s2 F4 = -5.0 N Notes

Example #1 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°? 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 2 kg object +8.0 N -Fg +FN F = +8.0 N +8.0 N 0 N frictionless Net force = +8.0 N a = 4.0 m/s2 Not in notes

Determining Net Force – Side View Consider each dimension separately. On a flat surface total vertical force = 0. Hor. Vert. 2 kg object +8.0 N -2.0 N -Fg +FN Ff = -2.0 N F = +8.0 N +6.0 N 0 N Net force = +6.0 N a = 3.0 m/s2 Notes

Example #2 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?

Determining Net Force – Vertical Object moving straight up or down  total horizontal force = 0. Hor. Vert. 20 N object in freefall -20 N 0 N -20 N Fg = -20 N Net force = -20 N a = -9.81 m/s2 Not in notes

Example #3 What is the net force acting on a 3000 newton rocket if its engine produces an upward thrust of 4500 newtons? Fthrust 1500 N upward Fg

End of 2.1.2A - PRACTICE