Physics 11
Fall 2012
NAME:
Quiz 6 - Solutions
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be giving partial credit, be sure to attempt all the problems, even if you don’t finish them!
A 5.0 kg cat and a 2.0 kg bowl of tuna fish are at opposite ends of a 4.0 m long seesaw.
How far to the left of the pivot must a 4.0 kg cat stand to keep the seesaw balanced?
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12.34.
Model:
Visualize:
The see-saw is a rigid body. The cats and bowl are particles.
Solution
This is just a statics problem since none of
the components of the system is moving.
The bowl and the smaller cat will tend to
twist the seesaw one way, while the heavier
cat will twist it the other way. Thus, the
seesaw is subject to three torques. If the
system is to balance, then
P the sum of the
torques must be zero,
τi = τnet = 0.
So, Solve:
we just need
to add is
upinthe
differentequilibrium.
torques, noting
that the
the net
bowl
and about
smaller
The see-saw
rotational
Calculate
torque
thecat
pivot point.
will lead to a positive torque, while the bigger cat will give a negative torque. So,
τ net = 0 = ( FG )1 ( 2.0 m ) − ( FG )2 ( d ) − ( FG ) B ( 2.0 m )
τnet = τbowl + τsmall cat − τbig cat = 0.
m2 gd = m1 g ( 2.0 m ) − mB g ( 2.0 m )
Now, in general, τ = rF sin φ. The force each component is just its mass times gravity.
( m1 − mand
( 5.0 kg − 2.0 kg )( 2.0 m ) 1.5 m
B )( 2.0 m )
Since gravity is always pointing straight
since the
d = down,
= seesaw is horizontal, the =angle
m2
4.0 kg
of each torque is 90◦ . Thus, we have
Assess:τnet The
close
allτbig
thecatway to the end= by0.the bowl, which makes sen
= smaller cat isτbowl
+ τbut
smallnot
cat −
combined mass
ofbowl
therbowl
smaller
cat
andcatbowl
tuna
isbiggreater
= m
g+m
rsmall of
−m
= mass
0. of the larger cat.
small
cat g
cat rbig than
cat g the
So, we just need to figure out rsmall cat , since all the rest are known. Solving for this gives
rsmall cat =
mbig cat rbig cat − mbowl rbowl
5(2) − 2(2)
=
= 1.5 m.
msmall cat
4
So, the small cat is pretty close to the bowl, but still a half a meter away.
1