Download Lecture 11

Center of mass and center of gravity
In a uniform gravitational field, the center of gravity is the
same as the center of mass.
It is as though all the weight of an object is concentrated
in a single point.
Two conditions of equilibrium
ìTotal force = 0
ì
ìTotal torque = 0
Or equivalently:
Reason:
Non-zero force will cause the center of mass to accelerate
Non-zero torque will cause the object to rotate
The torque can be calculated about any point as long as
the first condition is satisfied. If the total torque is zero
about one point, it will be zero about any other point (if
the first condition is satisfied).
Example
F1
What are F1 and F2?
F2
1m
2m
Total Force = 0N
Þ F1 + F2 - 300N = 0N
Now need one more equation.
F3=300N
Example
F1
P3
F2
1m
P1
P2
2m
F3=300N
Total Torque = 0Nm
But where is the pivotal point?
Answer: It doesn't matter! Pick any point you want.
Pick P1
F1
F2
P3
1m
P1
P2
2m
Suppose I pick P1 as the pivital point.
The total torque about P1 :
(3m)F2 - 2m(300N) = 0Nm
Þ F2 = 200N
Now use F1 + F2 - 300N = 0N
Þ F1 = 100N
F3=300N
Pick P2
F1
F2
P3
1m
P1
P2
2m
Suppose I pick P2 as the pivital point.
The total torque about P2 :
1m(300N) - (3m)F1 = 0Nm
Þ F1 = 100N
Now use F1 + F2 - 300N = 0N
Þ F2 = 200N
F3=300N
Pick P3
F1
F2
P3
1m
P1
P2
2m
F3=300N
Suppose I pick P3 as the pivital point.
The total torque about P3 :
(1m)F2 - 2mF1 = 0Nm
Now use F1 + F2 - 300N = 0N
Þ F2 = 2F1
Þ F1 + 2F1 - 300N = 0N
Þ F1 = 100N
Þ F2 = 200N
A rescue
Find n2, fs, and the coefficient of friction μ.
Use the conditions
Vertical force = 0 :
n2 - 180 - 800 = 0 Þ n2 = 980N
Horizontal force = 0 :
f s - n1 = 0 Þ f s = n1
Torque about B = 0 :
n1 (4m) - (180N)(1.5m) - (800N)(1m) = 0
Þ n1 = 268N
Put back n1 back into the second equation :
Þ f s = n1 = 268N
f s = mn2
Þm=
fs
= 0.27
n2
The torque could have been evaluated at any other points.
Mass: m
Example
Length: L
Wall: frictionless
F2
Ground: coefficient of friction m
What is the minimum angle f possible?
What are the forces?
mg
F1
φ
μF1
Total Force
F2
Horizontal total force:
Fx = m F1 - F2
Vertical total force:
Fy = F1 - mg
In equilibrium, both components must vanish:
mg
ìm F1 - F2 = 0
ì
ì F1 - mg = 0
ìF1 = mg
ì ì
ìF2 = m F1 = m mg
F1
φ
μF1
Total Torque
F2
I choose to pick the bottom as the pivotal point
(You can pick any other point you like)
L
Total torque:
1
t = L sin f F2 - ( L cos f )mg
2
1
= mgL( m sin f - cos f ) º 0
2
1
m sin f - cos f = 0
2
sin f
1
Þ tan f =
=
cos f 2 m
L sinϕ
L/2
mg
F1
φ
μF1
(L/2)cosϕ