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 free-fall

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 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.