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

2. 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.”

3. (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

4. 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 = (“upwards” = “downwards”)
2Tsin37 = 500
T = 500/(2sin37)
= 415 N

5. Brain Break!

6. 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

7. 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!