1. Drawing Pictures II
Forces on Angles

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

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

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

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

6. 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

7. Equilibrants – Top View
Resultant

8. Equilibrants – Top View
Resultant
Equilibrant

9. Equilibrants – Top View
Resultant

10. 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

11. 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

12. 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.

13. 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.

14. End