Transcript Forces & FBD's
Drawing Pictures II
2.1.2B 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
Equilibrant Resultant
Equilibrants – Top View
Resultant Equilibrant
Equilibrants – Top View
Resultant Equilibrant
Forces on Angles – Side View
• Determine net horizontal force. • Vertical forces must total zero. • F
g = F Y + F N …. F N < F g F F N F Y F X F g
H F X V +F Y -F g +F N F X 0 N
Net force = F X a = F X /m
Forces on Angles – Side View
• Determine net horizontal force. • Vertical forces must total zero. • F
N = F Y + F g …. F N > F g F N F g F Y F F X
H F X V -F Y -F g +F N F X 0 N
Net force = F X a = F X /m
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?
F X = F cos θ = 20N cos 30° = 17.3 N
– What is the acceleration of the box?
a = F net /m a = 17.3 N / 40 kg a = 0.43 m/s 2
– How could this acceleration be increased?
Decrease the angle of the pull or increase 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?
F X = F cos θ = 15N cos -60° = 7.5 N
– What is the acceleration of the mower?
a = F net /m a = 7.5 N / 30 kg a = 0.25 m/s 2
– How could this acceleration be decreased?
Increase the angle of the pull or decrease force.
End of 2.1.2B - PRACTICE