Coverage of the 1st Long test

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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
37°
37°
37°
Static Equilibrium
Condition 1
Σ Fx = 0
Σ Fy = 0
Σ Fz = 0
Condition 2
The sum of the
torque, ז, is zero.
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.
Static
60° 60°
Object in freefall
Fgravity
Objects falling at constant velocity (terminal velocity)
F air friction
F gravity
Object sliding at constant velocity(the surface is frictionless)
F normal
F gravity
Object sliding without friction
F gravity
F normal
Is the box accelerating?
mass of the block = 10kg
10N 5N
F applied Ff F normal
10 N
5N
F gravity
Is the box accelerating? mass of the block = 10kg
Σ Fy = Fnormal + Fgravity= 0
Fgravity = Weight = mass X acceleration = 10 kg X  9.8 m/s2
=  98 kgm/s2
=  98 N
Σ Fy = Fnormal + ( 98 N) = 0 Fnormal = 98 N
F normal
F applied = 10 N
Ff = 5N
F gravity
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
Note: There is friction between the load and the incline.
• http://www.google.com.ph/imgres
Draw the FBD of the box
Draw the FBD of the knot (include the angle).
A B
C
Summary
• The object is in static equilibrium if it is not moving and not rotating.
• A freebody 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.