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4. We note that
ma
→
= (–16 N) i
^
+ (12 N) j
^
.
With the other forces as specified in the
problem, then Newton’s second law gives the third force as
F
3
→
=
→
–
F
1
→
–
F
2
→
=(–34 N) i
^
−
(12 N) j
^
.
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View Full Document 32. We resolve this horizontal force into appropriate components.
(a) Newton’s second law applied to the
x
axis
produces
Fm
g
m
a
cos
sin
.
θ
−=
For
a
= 0, this yields
F
= 566 N.
(b) Applying Newton’s second law to the
y
axis (where there is no acceleration), we have
sin
cos
0
N
FF
m
g
θθ
−−
=
which yields the normal force
F
N
= 1.13
×
10
3
N.
mg
F
w
sin
.
θ
−=
0
This yields
F
w
= 68 N (uphill).
(b) Given our coordinate choice, we have
a
=
a
= 1.0 m/s
2
. Newton’s second law
mg
F
ma
w
now leads to
F
w
= 28 N (uphill).
(c) Continuing with the forces as shown in our figure, the equation
mg
F
ma
w
will lead to
F
w
= – 12 N when 
a
 = 2.0 m/s
2
. This simply tells us that the wind is
opposite to the direction shown in our sketch; in other words,
12 N
w
F
=
G
downhill
.
36. We label the 40 kg skier “
m
” which is represented as a block in the figure shown. The
force of the wind is denoted
G
F
w
and might be either “uphill” or “downhill”
(it is shown
uphill in our sketch). The incline angle
is 10°. The
−
x
direction is downhill.
(a) Constant velocity implies zero acceleration; thus, application of Newton’s second law
along the
x
axis leads to
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View Full Document50. First, we consider all the penguins (1 through 4, counting left to right) as one system,
to which we apply Newton’s second law:
()
(
)
4
1234
2
222N
12kg
15kg
20kg
.
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This note was uploaded on 11/16/2009 for the course PHYSICS 11a taught by Professor Silvera during the Fall '07 term at Harvard.
 Fall '07
 silvera
 mechanics, Force

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