Physics: Free Body Diagram

General Method for Solving Force Problems

  1. Identify the object(s) of interest
  2. Draw a separate free body diagram for each object
  3. For each diagram, do the following:
    1. Draw all forces on the free body as vectors coming out of the object
    2. Draw all other information (velocity, acceleration, etc.) away from the object
    3. Label an appropriate coordinate axis
    4. Identify and Label the x- and y- components of all forces
    5. Write Newton's Second Law in component form:
      SFx=max,
      SFy=may,
    6. Find SFx and SFy from the diagram
    7. Find max and may from your knowledge of the problem
  4. Write any other relevant equations, such as for friction, centripetal acceleration, etc.
  5. Solve for your answer

Notes to Remember:

  1. Pay attention to sign conventions, based on your choice of axes
  2. Only draw forces on the object itself, not velocities or accelerations
  3. Only draw forces that act on the object
  4. Any object "at rest" or "at constant velocity" in any dimension (x or y) has that component of its acceleration (ax or ay) equal to zero
  5. Normal means "perpendicular to a surface". There is a normal force only if an object is against a surface, and it always points away from that surface
  6. Tension exists only if there is a rope/string pulling on an object. Tension always points along the rope/string, and it is the same at both ends.
    (this is important if you have a rope attached to two objects, as in a pulley problem)
  7. The centripetal force is not an extra force on the diagram. It is just a name for the net force in the center-pointing direction.

Example:

What angle of inclination q of the ramp shown will cause a block of mass m (already in motion) to slide with constant velocity? The coefficient of kinetic friction between the block and the ramp is mk. How would your answer change for a block of mass 2m?

Solution:
 

"Constant velocity" means that ax = 0.
Since the block is not burrowing into
the ramp, not lifting off from it,
ay = 0, as well.

Kinetic friction (since in motion):
    Ff = mkFN

Components of Fg:
    Fgx = Fg sin q
    Fgy = Fg cos q

Newton's Second Law (in component form):

                        SFx = max
(Note the signs)      Fgx - Ff = 0

(Substitute)    Fgsin q - mkFN= 0
(Solve)         Fg sin q= mkFN

                       SFy = max
(Note the signs)     FN - Fgy = 0

(Substitute)    FN - Fgcos q = 0
(Solve) FN = Fgcos q

(Combine the two solutions
in bold above)

Fgsin q = mkFgcos q
sin q = mkcos q
tan q = mk
q = tan-1mk

This answer is independent of the mass of the block

















 

Page developed by CAPS Tutor, Nick Menicucci 10/23/01