Transcript Static Equilibrium (print version)

Consider a meter stick of mass 0.306 kg pivoted at one end which is held
horizontally by a vertical string.
a. Draw an extended force diagram of the meter stick.

Fg
What forces act upon the meter stick?
Where should the force of gravity be considered to act?
Center of Gravity – the single point on a body where the force of gravity
effectively acts.
As long as the local gravitational field strength (g) is essentially constant, the center of
gravity is the same as the center of mass.
Consider a meter stick of mass 0.306 kg pivoted at one end which is held
horizontally by a vertical string.
a. Draw an extended force diagram of the meter stick.

FT

FN

Fg
Do all three force exert torques around the pivot?
The Normal Force does not exert a
torque because it acts radially
(through the axis of rotation)!
Consider a meter stick of mass 0.306 kg pivoted at one end which is held
horizontally by a vertical string.
b. Complete the torque bar graph quantitatively.

FT

FN
 N  g T
 net

Fg
 g  r Fg
 g  r mg
0
 g  0.50 m0.306 kg  9.8 N kg 

 g  1.5 Nm
-1.5 Nm

Consider a meter stick of mass 0.306 kg pivoted at one end which is held
horizontally by a vertical string.
b. Complete the torque bar graph quantitatively.

FT

FN
1.5 Nm
 N  g T
 net

Fg
Since the angular velocity of the meter
stick doesn’t change, the torques must
be balanced!
0
 net  ?0
Newton’s First Law!
-1.5 Nm
Consider a meter stick of mass 0.306 kg pivoted at one end which is held
horizontally by a vertical string.
c. What is the tension in the string?

FT

FN
1.5 Nm
 N  g T
 net

Fg
 T  1.5 Nm
r FT  1.5 Nm
0
-1.5 Nm
1.5 Nm
FT 
r
1.5 Nm
FT 
1.0 m
FT  1.5 N
Consider a meter stick of mass 0.25 kg pivoted at one end which is held
horizontally by a vertical string.
Equilibrium

Occurs when  F  0,   0

FT

FN
Static Equilibrium

Occurs when  F  0,   0, and
v   0

Fg
FN  FT  Fg Since v  constant
FN  Fg  FT
FN  mg  FT
FN  0.306 kg  9.8 N   1.5 N
kg 

3.0 N 
FN  1.5 N