Chapter 6
167
*P6.22
We adopt the view of an inertial observer. If it is on the verge of sliding, the
cup is moving on a circle with its centripetal acceleration caused by friction.
Fm
a nm
g
a f
mv
r
nm
g
yy
xx
ss
∑
∑
=+
−
=
==
=
=
:
:
0
2
µµ
vg
r
s
=
µ
0898
30
153
..
.
ms
m
2
ej
af
mg
f
n
FIG. P6.22
If you go too fast the cup will begin sliding
straight across the dashboard to the left.
P6.23
The only forces acting on the suspended object are the force of gravity
m
g
and the force of tension
T
, as shown in the freebody diagram. Applying
Newton’s second law in the
x
and
y
directions,
FT
m
a
x
∑
sin
θ
(1)
m
g
y
∑
=−
=
cos
0
or
Tm
g
cos
=
(2)
T
cos
T
sin
mg
FIG. P6.23
(a)
Dividing equation (1) by (2) gives
tan
.
.
.
=
a
g
300
980
0306
±ms
2
2
.
Solving for
,
=°
17 0
.
(b)
From Equation (1),
T
ma
°
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics, Acceleration, Friction, Inertia

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