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Applying Newton&s Laws
533
Dividing the two equations:
slope
string
slope
sin
tan
cos
ag
θ
θ
g
θ
+
=
For the crate, the component of the weight along the slope is
cs
l
o
p
e
sin
mg
θ
−
and the normal force is
l
o
p
e
cos
.
θ
Using Newton&s Second Law again:
c
slope
k
c
slope
c
sin
cos
θ
θ
ma
μ
−+
=
.
slope
k
slope
sin
cos
θ
g
θ
+
=
. This leads to the
interesting observation that the string will hang at an angle whose tangent is equal to the coefficient of kinetic
friction
:
k
string
tan
tan(90
68 )
tan 22
0.40
θ
==
°
−
°
=
°
=
.
EVALUATE:
In the limit that
k
0
→
,
string
0
→
and the string is perpendicular to the top of the crate.
As
k
increases,
string
increases.
Figure 5.75
5.76.
IDENTIFY:
Apply
m
∑
F
=a
!
!
to yourself and calculate
a
. Then use constant acceleration equations to describe
the motion.
SET UP:
The freebody diagram is given in Figure 5.76.
EXECUTE:
(a)
y
y
Fm
a
∑=
gives
cos
nm
g
α
=
.
x
x
a
gives
k
sin
mg
α
fm
a
−=
. Combining these two
equations, we have
2
k
(sin
cos )
3.094 m s
αμ
=−
=
−
. Find your stopping distance:
2
0
0,
3.094 m s ,
20 m s
xx
x
va
v
−
=
.
22
000
2
(
) gives
64.6 m,
x
vv a
=+
−
which is greater than 40 m.
You don&t stop before you reach the hole, so you fall into it.
(b)
2
0
3.094 m s ,
40 m,
0
ax
x
v
−
=
=
.
00
0
2
(
) gives
16 m s.
x
x
v
−
=
EVALUATE:
Your stopping distance is proportional to the square of your initial speed, so your initial speed is
proportional to the square root of your stopping distance. To stop in 40 m instead of 64.6 m your initial speed must
be
40 m
(20 m/s)
16 m/s
64.6 m
=
.
Figure 5.76
5.77.
IDENTIFY:
Apply
m
∑
F
!
!
to each block and to the rope. The key idea in solving this problem is to recognize
that if the system is accelerating, the tension that block
A
exerts on the rope is different from the tension that block
B
exerts on the rope. (Otherwise the net force on the rope would be zero, and the rope couldn&t accelerate.)
SET UP:
Take a positive coordinate direction for each object to be in the direction of the acceleration of that
object. All three objects have the same magnitude of acceleration.
EXECUTE:
The Second Law equations for the three different parts of the system are
:
Block
A
(The only horizontal forces on
A
are tension to the right, and friction to the left):
k
.
AA
A
mg T
=
Block
B
(The only vertical forces on
B
are gravity down, and tension up):
.
BBB
Rope (The forces on the rope along the direction of its motion are the tensions at either end and the weight of the
portion of the rope that hangs vertically):
( )
.
RB
A
R
d
T
T
m
a
L
+−=
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View Full Document534
Chapter 5
To solve for
a
and eliminate the tensions, add the left hand sides and right hand sides of the three equations
:
( )
k
k
(/)
()
,
o
r
.
B
RA
ABR
A
B
R
mm
d
L m
d
mg mg m
g
m
m
m a
a g
L
mmm
μ
+−
−+
+
=
+
+
=
++
(a)
When
k
0,
.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Force

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