Making Pretty Pictures Free Body Diagrams and Net Force

Free Body Diagrams VECTOR diagrams! Shows ALL FORCES acting on an object Must be properly labeled Motionless Equilibrium FN Fg

unbalanced force, it is also Fnet Not moving up or down, so… Net Force X Y Net Force = 0 *Equilibrium* Motionless or Constant Velocity Net Force ≠ 0 Accelerating or Decelerating FA Fg FN Total = FA Total = 0 Total = FA Since this is the only unbalanced force, it is also Fnet Not moving up or down, so… Fg + FN = 0 Fg = -FN FN m Fg frictionless

Example #1 A man pushes a 50 kilogram crate across a frictionless surface with a constant force of 100 Newtons. What is the normal force that pushes on the crate? Draw a free-body diagram of the crate. What is the net force on the crate? What is the crate’s acceleration? What is the weight of the crate? Fnet will only be the 100N horizontal force FA FN Fg Fg = mg Fg = (50 kg)(9.81 m/s2) Fg = 490.5 N FN = Fg FN = 490.5 N a = Fnet / m a = (100 N) / (50 kg) a = 2 m/s2

Example #2 A horse pulls a 500 kilogram sled with a constant force of 3,000 Newtons. The force of friction between the sled and the ground is 500 Newtons. What is the normal force that pushes on the sled? Draw a free-body diagram of the sled. What is the net force on the sled? What is the sled’s acceleration? What is the weight of the sled? Fnet = ΣFx Fnet = 3000 N – 500 N Fnet = 2500 N Fg = mg Fg = (500 kg)(9.81 m/s2) Fg = 4905 N FA FN Fg Ff FN = Fg FN = 4905 N a = Fnet / m a = (2500 N) / (500 kg) a = 5 m/s2

This applied force (FA) The total vertical force must Pulling on an Angle A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of 30° above horizontal. This applied force (FA) can be broken into COMPONENTS X Y FAX FAY Fg FN Total = FAX Total = 0 FN The total vertical force must be 0, so FN = Fg - FAY FA FAY 30˚ FAX Acceleration depends only on FAX Fg

Example #3 A man pulls a 40 kilogram crate across a smooth, frictionless floor with a force of 20 N that is 45˚ above horizontal. What is the net force on the sled? How could the acceleration be increased? Fnet = FA cos θ Fnet = (20 N)(cos 45°) Fnet = 14.14 N Pushing at a smaller angle will make Fnet greater and therefore increase acceleration. What is the crate’s acceleration? a = Fnet / m a = (14.14 N) / (40 kg) a = 0.35 m/s2

This applied force (FA) The total vertical force must Pushing on an Angle A block is pushed along a frictionless, horizontal surface with a force of 100 newtons at an angle of 30° below horizontal. This applied force (FA) can be broken into COMPONENTS X Y FAX FAY Fg FN Total = FAX Total = 0 FN The total vertical force must be 0, so FN = Fg + FAY FAX FAY -30˚ FA Acceleration depends only on FAX Fg

Example #4 A girl pushes a 30 kilogram lawnmower with a force of 15 Newtons at an angle of 60˚ below horizontal. Assuming there is no friction, what is the acceleration of the lawnmower? Fnet = FA cos θ Fnet = (15 N)(cos 60°) Fnet = 7.5 N a = Fnet / m a = (7.5 N) / (30 kg) a = 0.25 m/s2 What could she do to reduce her acceleration? Push at an greater angle