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 normal F gravity

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

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

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.