1. Coverage of the 1st Long test
Newton’s Laws of Motion
graphing vectors
calculating components of vectors
(horizontal and vertical components)
getting the resultant vector using the component method
free- body diagram
Static equilibrium
calculating forces involved in a given system

2. 37°

3. 37°

4. 37°

5. Static Equilibrium Condition 1 Σ Fx = 0 Σ Fy = 0 Σ Fz = 0 Condition 2
The sum of the
torque, ז, is zero.

9. Center of Gravity the point where the force of gravity is concentrated
When the center of
gravity falls within the
base of the object, then
the object is stable.

12. Static
60°
60°

13. Object in free-fall
Fgravity

14. Objects falling at constant velocity (terminal velocity)
F air friction
F gravity

15. Object sliding at constant velocity (the surface is frictionless)
F normal
F gravity

16. Object sliding without friction
F normal
F gravity

17. Is the box accelerating?
mass of the block = 10kg
10N N
F applied Ff
F normal
10 N 5N
F gravity

18. Is the box accelerating?
mass of the block = 10kg
Σ Fy = Fnormal + Fgravity= 0
Fgravity = Weight
= mass X acceleration
= 10 kg X m/s2
= kgm/s2
= - 98 N
Σ Fy = Fnormal + ( -98 N) = 0
Fnormal = 98 N
F normal
F applied = 10 N
Ff = 5N
F gravity

19. Is the box accelerating?
mass of the block = 10kg
Σ F x = Fapplied + Ff
Σ Fx = 10 N + (-5N) = 5N
Fnet = 5N = massX acceleration
5N = 10 kg X a
a = 0.5 m/s2
F normal
F applied = 10 N
Ff = 5N
F gravity

20. Note: There is friction between the load and the incline.
Draw the FBD of the box

21. A B C
Draw the FBD of the knot (include the angle).

22. Summary
The object is in static equilibrium if it is not moving and not rotating.
A free-body diagram can be drawn to evaluate the effect of forces on the object.
There is always a force of gravity (also known as weight) which is equal to the product of the mass and acceleration due to gravity.
There is a normal force perpendicular to the surface that supports and balances the object vertically.