Biomedical Engineering Reference
In-Depth Information
2.5 Rotational Equilibrium
For a body to be in rotational equilibrium, the sum of the torques produced by all the
forces acting on the body must be zero. This is the condition used by the two-pan
balance. The torque of the weight of an object in one of the pans of the balance
makes it fall (rotate), if there is nothing in the other pan. As the calibrated mass is
placed on the other pan, the torque due to its weight begins to equilibrate the
balance. When the weights on both pans are the same, the magnitude of both
torques will be the same and the net torque will be zero, due to opposite signs,
and the balance will be in equilibrium.
Example 2.3 John and Mary are playing on a seesaw, as can be seen in the figure of
Example 2.3. John, with a mass of 20 kg, is seated 2 m from the pivot point, called
the fulcrum. At what distance from the fulcrum Mary, with a mass of 30 kg, has to
sit so that the board will be in horizontal equilibrium?
Mary
John
d M
d J
The sketch shows the situation:
To be in equilibrium, the sum of torques of weights of Mary and John must be
zero, that is,
M M þ
M J ¼
0,
2m
s 2
M J ¼
ð
Johns mass
Þ
gd J ¼
ð
20kg
Þ
10m
=
ðÞ¼
400Nm
:
(30 kg)(10 m/s 2 ) d M .
Hence, in equilibrium, M M ¼
+ 400 N m
¼
(400 N m)/(300 kg m/s 2 )
1.33 m.
As Mary weighs more than John, she has to sit closer to the fulcrum of the
seesaw than John for the seesaw to be in horizontal equilibrium, that is, at 1.33 m
from the pivot point O.
Then, d M ¼
¼
Exercise 2.6 You want to construct a mobile with four ornaments and three light
rods with negligible mass as shown in the figure of Exercise 2.6. The distances in
centimeters and the mass of one of the ornaments in grams are indicated in the
figure. Determine the masses of ornaments A, B, and C, considering that the mobile
should be in equilibrium.
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