What’s the lever arm between F and the fulcrum?
A. l
B. a
C. b
a
b
When the applied force is not perpendicular to the
crowbar, for example, the lever arm is found by drawing
the perpendicular line from the fulcrum to the line of
action of the force.
We call torques
that produce
counterclockwise
rotation positive,
and torques that
produce
clockwise rotation
negative.
Two forces are applied to a merry-go-round with a
radius of 1.2 m as shown. What is the torque about
the axle of the merry-go-round due to the 80-N
force?
a)
b)
c)
d)
+9.6 N·m
-36 N·m
+96 N·m
-36 N·m
1.2 m  80 N = +96 N·m
(counterclockwise)
What is the torque about the axle of the
merry-go-round due to the 50-N force?
a)
b)
c)
d)
+60 N·m
-60 N·m
+120 N·m
-120 N·m
-(1.2 m  50 N) = -60 N·m
(clockwise)
What is the net torque acting on the merrygo-round?
a)
b)
c)
d)
e)
+36 N·m
-36 N·m
+96 N·m
-60 N·m
+126 N·m
96 N·m (counterclockwise)
- 60 N·m (clockwise)
= +36 N·m (counterclockwise)
How far do we have to place the 3-N
weight from the fulcrum to balance the
system?
a)
b)
c)
d)
2 cm
27 cm
33 cm
53 cm
F=3N
 = +1 N·m
l=/F
= (+1 N·m) / (3 N)
= 0.33 m = 33 cm
The center of gravity
is the point about which the weight of the object itself
exerts no torque.
We can locate the center of gravity by finding the point
where it balances on a fulcrum.
For a more complex object, we locate the center of gravity by
suspending the
object from two
different points,
drawing a line straight
down from the point of
suspension in each
case, and locating the
point of intersection of
the two lines.
1J-21 Center of Gravity of an Irregular Lamina
How can we find the Center of Gravity of Irregular shapes?
Why does the mass
on the string hang
straight down ?
Where is the Center
of Gravity of the cut
board ?
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Quiz: where is the CG of the beam and the CG of the
system (beam + two weights) after it’s balanced?
a)
b)
c)
CG of the beam is Center of the beam, CG of the
system is at the pivot point.
Both are at the center of the beam.
Both are the pivot points.
If the center of gravity
lies below the pivot
point, the object will
automatically regain
its balance when
disturbed.
The center of gravity
returns to the position
directly below the
pivot point, where the
weight of the object
produces no torque.
1J-23 Corks & Forks
Can the Center of Gravity lie at a point not on the object?
How difficult is it to
balance this system on
a sharp point ?
Where is the C
of G ?
THE CENTER OF GRAVITY IS NOT ON THE OBJECT. IT ACTUALLY LIES ALONG THE
VERTICAL BELOW THE SHARP POINT.
WHEN THE FORK IS MOVED THE CM RISES AND THIS MEANS THE SYSTEM IS IN STABLE
EQUILIBRIUM .
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Fall
2009
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1J - 24 Climbing double cone
What happens to the
center of mass ?
A). Going down the hill
B). Going uphill
C). Stay at rest
D). Depend on the object shape
The force causing the object to move is gravity and we know that by energy
conservation that if the object gains kinetic energy it must lose potential energy.
Therefore the center of mass must be falling and the kinetic energy = mgh where h
is the distance the CM falls.
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How far can
the child walk
without tipping
the plank?
• For a uniform plank, its center of gravity is at its geometric center.
• The pivot point will be the edge of the supporting platform.
• The plank will not tip as long as the counterclockwise torque from the
weight of the plank is larger than the clockwise torque from the
weight of the child.
• The plank will verge on tipping when the magnitude of the torque of
the child equals that of the plank.
1J-16 Walk the Plank
What happens when a mass is placed at the end of a massive plank?
Sum of Torque about Pivot
XMg–xmg=0
m=MX/x
Can you safely
walk to the
end of the
plank ?
One can solve for either M or m, if
the other quantity is known
EVEN WITH A MASS AT THE END OF THE PLANK, THE SYSTEM CAN STILL BE IN
EQUILIBRIUM
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Fall
2009
Fall 2009
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1J-28 Wine Bottle Holder
Balance a Bottle and a Wooden Holder by Eliminating Net Torque
M
How does this
system Balance?
m1
x1
x2
m2
Sum of Torque about Pivot
m1x1g - m2x2g = 0
THE CENTER OF GRAVITY OF THE BOTTLE PLUS THE WOOD MUST LIE DIRECTLY OVER AND
WITHIN THE BOUNDARY OF THE SUPPORT (PIVOT). FOR BALANCE THERE CAN BE NO
NET TORQUE ON SYSTEM.
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Fall
2009
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1J-29 Overhanging Blocks
How far out from the table can a stack of bricks be balanced ?
Length = L, CG = L/2
Length = 3L/2, CG = 3L/4
Length = 7L/4, CG = 25L/24
How do the blocks
stay balanced when
the top block extends
beyond the bottom
block ?
Δx
Blocks can over hang but the Center of Gravity
of a block must be inside the block below
For 6 blocks max extension Δx :
Δx = L/2 + L/4 + L/6 + L/8 + L/10 + L/12 = 1.22(L)
IN REALITY EACH BLOCK HAS TO BE MOVED SLIGHTLY BACK TO AVOID TIPPING, SO
THE TOTAL EXTENSION WILL BE A LITTLE LESS.
2/23/2011
An 80-N plank is placed on a dock as shown. The
plank is uniform in density so the center of gravity of
the plank is located at the center of the plank. A
150-N boy standing on the plank walks out slowly
from the edge of the dock. What is the torque
exerted by the weight of the plank about the pivot
point at the edge of the dock?
a)
b)
c)
d)
e)
f)
+80 N·m
-80 N·m
+160 N·m
-160 N·m
+240 N·m
-240 N·m
1 m  80 N = +80 N·m
(counterclockwise)
Quiz: How far from the edge of the dock can the
150-N boy walk until the plank is just on the verge of
tipping?
a)
b)
c)
d)
0.12 m
0.23 m
0.53m
1.20 m
80 N·m / 150 N = 0.53 m