x
x
2
2
x
F
ma
The only force acting on the crate is friction.
f
N
(mg)
ma
a
g
(0.250)(9.8m /s )
2.45m /s
=
= μ
= μ
=
= μ
=
=
∑
!
!
nd
A
B
First, we assume a direction of motion, which for this problem would be
the ydirection, and let it be positive. Let's apply Newton's 2
Law to each block
individually:
Block A: T = m a
Block B: m g
T

B
B
B
A
B
B
A
m a
Because the blocks are connected, the acceleration must be the same. Notice that
we have +T and T in the 2 equations. If we add them together, T will be eliminated.
m g
a(m
m )
m g
(1.60
a
(m
m )
=
=
+
=
=
+
2
2
kg)(9.8m / s )
3.14m / s
(3.40
1.60)kg
=
+
f
N
= μ
!
!
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View Full DocumentThe apparent weight is the Normal force.
2
2
2
2
F
ma
Everything is happening in the
ydirection, so we can
drop the vector notation.
v
N
mg
m
Substitute 7mg for the Normal force.
r
v
7mg
mg
m
Notice that the pilot's mass cancels out.
r
v
(320m / s)
r
8g
6
=
±

=

=
=
=
∑
!
!
2
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 Fall '08
 Ashkenkai
 Physics, Force, Friction, Normal Force, hard rubber ball

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