Drawing Pictures II Forces on Angles

Forces on Angles – Top View Redraw force vectors head-to-tail then draw resultant from start to finish. Resultant

Forces on Angles – Top View Redraw force vectors head-to-tail then draw resultant from start to finish. Resultant

Forces on Angles – Top View Redraw force vectors head-to-tail then draw resultant from start to finish. Resultant

Forces on Angles – Top View Redraw force vectors head-to-tail then draw resultant from start to finish. Resultant

Equilibrants – Top View Equilibrant is the force that is added to a system to produce a net force of ZERO. Opposite of the RESULTANT Equilibrant Resultant

Equilibrants – Top View Resultant

Equilibrants – Top View Resultant Equilibrant

Equilibrants – Top View Resultant

Forces on Angles – Side View Determine net horizontal force. Vertical forces must total zero. Fg = FY + FN …. FN < Fg F H V FY FN FX +FY -Fg +FN FX FX 0 N Fg Net force = FX a = FX/m

Forces on Angles – Side View Determine net horizontal force. Vertical forces must total zero. FN = FY + Fg …. FN > Fg H V FN FX -FY -Fg +FN FX FX 0 N FY Net force = FX a = FX/m Fg F

Decrease the angle of the pull or increase force. Example #1 A 40 kilogram box is pulled across a smooth, frictionless surface with a 20 newton force that is 30° above horizontal. What is the net force acting on the box? What is the acceleration of the box? How could this acceleration be increased? FX = F cos θ = 20N cos 30° = 17.3 N a = Fnet/m a = 17.3 N / 40 kg a = 0.43 m/s2 Decrease the angle of the pull or increase force.

Increase the angle of the pull or decrease force. Example #2 A woman pushes a 30 kilogram lawnmower with a force of 15 newtons at an angle of 60° below horizontal. What is the net force acting on the mower? What is the acceleration of the mower? How could this acceleration be decreased? FX = F cos θ = 15N cos -60° = 7.5 N a = Fnet/m a = 7.5 N / 30 kg a = 0.25 m/s2 Increase the angle of the pull or decrease force.

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