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1
Chap. 12.
HW
Questions
2.
Figure 1216 shows four overhead views
of rotating uniform disks that are sliding
across a frictionless floor. Three forces, of
magnitude
F
,
2F
, or
3F
, act on each disk,
either at the rim, at the center, or halfway
between rim and center. The force vectors
rotate along with the disks, and, in the
“snapshots” of Figure 1216, point left or
right. Which disks are in equilibrium?
Assume positive direction is to RIGHT, assume positive rotation is counterclockwise [ccw]
a) sum of forces along horizontal = 0
sum of torques about center = FR – 2F R/2 +0 = 0
IN EQUILIBRIUM
b) sum of forces along horizontal = 4F
sum of torques about center = FR +0+FR
= 0
NOT IN EQUILIBRIUM
c) sum of forces along horizontal = 0
sum of torques about center = = FR +0+FR
= 0
IN EQUILIBRIUM
d) sum of forces along horizontal = 0
sum of torques about center = FR – 0  2FR = 3FR
NOT IN EQUILIBRIUM
3.
Figure 1217 shows three
situations
in
which
the
same
horizontal rod is supported by a hinge
on a wall at one end and a cord at its
other end. Without written calculation,
rank the situations according to the
magnitudes of (a) the force on the rod
from the cord, (b) the vertical force
on the rod from the hinge, and (c) the
horizontal force on the rod from the
hinge, greatest first.
a)
Take torque about hinge.
The torque due to
mg
must balance the torque due
T.
So for all
three cases the torque due to
T
must be the same.
This means that
T
is smallest for (2).
The
ranking is:
(1) and (3) tie for largest
T
then (2) is
smallest.
b) Taking torque about the right side eliminates
T
and only includes
mg
and
F
v
.
Since mg
does not change
F
v
is the same for all three cases.
c) If the
x
component of
T
is made bigger (i.e.
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 Spring '08
 oshea
 Force, Friction

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