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We calculate the torque around the step corner. The second diagram indicates that the
distance from the line of
F
to the corner is
r
–
h
, where
r
is the radius of the wheel and
h
is the height of the step.
The distance from the line of
mg
to the corner is
rr
h
r
h
h
2
2
2
2
+− =
−
b
g
. Thus,
Fr h mg rh h
−−
− =
bg
20
2
.
The solution for
F
is
22
2
2
2
2
2(6.00 10 m)(3.00 10 m) (3.00 10 m)
2
=
(0.800 kg)(9.80 m/s )
(6.00 10 m) (3.00 10 m)
13.6 N.
rh h
Fm
g
rh
−
××
−
×
−
=
−×
−
×
=
21. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force, Gravity

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