12.2 CHAPTER 12. FORCE,MOMENTUM AND IMPULSE
For the example, assume that the positive direction is to the right, then:
FR = (+4N) + (−6N)
=−2N
= 2N to the left
Remember that a negative answer means that the force acts in the opposite direction to the one that
you chose to be positive. You can choose the positive direction tobe any way you want, but once you
have chosen it you must keep it.
As you work with moreforce diagrams in whichthe forces exactly balance, you may notice thatyou
get a zero answer (e.g.0 N). This simply means that the forces are balanced and that the object will
not accelerate.
Once a force diagram has been drawn the techniques of vector additionintroduced in Chapter 11 can
be used. Depending onthe situation you might choose to use a graphicaltechnique such as the tail-to-
head method or the parallelogram method, or else an algebraic approach to determine the resultant.
Since force is a vector quantity all of these methods apply.
Example 3: Finding the resultant force
QUESTION
A car (mass 1200 kg) applies a force of 2000 N on a trailer (mass 250 kg). A constant frictional
force of 200 N is actingon the trailer, and a constant frictional force of 300 N is acting on the
car.
- Draw a force diagramof all the forces acting on the car.
- Draw a free body diagram of all the horizontal forces acting on the trailer.
- Use the force diagramto determine the resultant force on the trailer.
SOLUTION
Step 1 : Draw the force diagramfor the car.
The question asks us todraw all the forces on the car. This means that we must
include horizontal and vertical forces.
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FN: Upward force
Fg: Downward force of
Ff: Frictional force on car
F 1 : Force of trailer on car
(to the left) (2000 N)
(to the left) (300 N)
the Earth on car (12 000 N)
of road on car (12000 N)
Step 2 : Draw the free body diagram for the trailer.