Lesson 1: Newton’s First Law

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Presentation transcript:

Lesson 1: Newton’s First Law Unit 5: Equilibrium Lesson 1: Newton’s First Law

Newton’s First Law “An object remains in a state of uniform motion (constant velocity or rest) unless acted upon by a net unbalanced force.”

(force of gravity = mass x g) Example: A 600 kg spy-copter hovers silently in the night. Agent Glassford (100 kg) hangs from it, waiting to make his move Find the force exerted by the copter’s rotor blades. Step 1 : Draw a free body diagram (of the helicopter) R (rotor blade force) g = 9.8 N/kg (force of gravity = mass x g) mg (acrobat) (helicopter) Mg Step 2 : Balance the forces: [upward forces] = [downward forces] 10 min R = Mg + mg R =(600)(9.8) + (100)(9.8) R = 6860N

Two – Dimensional Problems A 500 N weight hangs as shown: Find the force exerted by each string. Step 1 : Draw a free body diagram y T1 T2 Y-component 37° 37° x X-component 500N 10 min Step 2 : Consider X and Y components of the forces separately. X : T1cos37 = T2cos37 (“left” forces = “right” forces) T1 = T2 (=T) Y : T1sin37 + T2sin37 = 500 (“upwards” = “downwards”) 2Tsin37 = 500 T = 500/(2sin37) = 415 N

Brain Break!

A 500 N weight is hung by two strings as shown. Your Turn! A 500 N weight is hung by two strings as shown. y TA TB 30º 60º x Find the tension in each string. TA TAsin30 30º TAcos30 500N X : TAcos30 = TBcos60 1 Y : (TA)sin30 + TBsin60 = 500 2 15 min TA = TB(cos60/cos30) 1 2 [ TB(cos60/cos30) ]sin30 + TBsin60 = 500 TB = 500/ [ (cos60•sin30/cos30) + sin60 ] = 433 N TA = 433(cos60/cos30) = 250 N 1

or y x X : 500sin30 = TA = 250 N Y : 500cos30 = TB = 433 N 30º x 500N 5 min X : 500sin30 = TA = 250 N Y : 500cos30 = TB = 433 N Try to line forces up with axes if possible!