homework 05 – FONTENOT, BRIAN – Due: Feb 18 2008, 4:00 am
1
Question 1, chap 5, sect 5.
part 1 of 2
10 points
A child holds a sled on a frictionless, snow
covered hill, inclined at an angle of 26
◦
.
87 N
μ
= 0
F
26
◦
If the sled weighs 87 N, find the force ex
erted on the rope by the child.
Correct answer:
38
.
1383
N (tolerance
±
1
%).
Explanation:
Given :
W
= 87 N
,
θ
= 26
◦
,
and
μ
= 0
.
Consider the free body diagram for the
block
m g
sin
θ
N
=
m g
cos
θ
μ
N
F
W
Basic Concepts:
If we ”tilt” our world,
and consider the forces parallel to the hill,
F
net
=
summationdisplay
F
up

summationdisplay
F
down
= 0
then the forces perpendicular to the hill,
F
net
=
summationdisplay
F
out

summationdisplay
F
in
= 0
Solution:
Consider the free body diagram
for the sled: The weight of the sled has compo
nents
W
sin
θ
acting down the hill and
W
cos
θ
acting straight into the hill.
The system is in equilibrium, so for forces
parallel to the hill,
F
net
=
T
 W
sin
θ
= 0
=
⇒
T
=
W
sin
θ
= (87 N) sin 26
◦
=
38
.
1383 N
.
Question 2, chap 5, sect 5.
part 2 of 2
10 points
What force is exerted on the sled by the
hill?
Correct answer:
78
.
1951
N (tolerance
±
1
%).
Explanation:
For forces perpendicular to the hill,
F
net
=
N  W
cos
θ
= 0
=
⇒ N
=
W
cos
θ
= (87 N) cos 26
◦
=
78
.
1951 N
.
Question 3, chap 5, sect 5.
part 1 of 2
10 points
Consider the 634 N weight held by two
cables shown below. The lefthand cable had
tension
T
2
and makes an angle of
θ
2
with
the ceiling. The righthand cable had tension
440 N and makes an angle of 36
◦
with the
ceiling.
The righthand cable makes an angle of 36
◦
with the ceiling and has a tension of 440 N
.
634 N
T
2
440 N
36
◦
θ
2
a) What is the tension
T
2
in the lefthand
cable slanted at an angle of
θ
2
with respect to
the wall?
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homework 05 – FONTENOT, BRIAN – Due: Feb 18 2008, 4:00 am
2
Correct answer:
517
.
319
N (tolerance
±
1
%).
Explanation:
Observe the freebody diagram below.
F
2
F
1
θ
1
θ
2
W
g
Note:
The sum of the
x
 and
y
components of
F
1
, F
2
,
and
W
g
are equal to zero.
Given :
W
g
= 634 N
,
F
1
= 440 N
,
θ
1
= 36
◦
,
and
θ
2
= 90
◦

θ .
BasicConcept:
Vertically and Horizontally,
we have
F
x
net
=
F
x
1

F
x
2
= 0
=
F
1
cos
θ
1

F
2
cos
θ
2
= 0
(1)
F
y
net
=
F
y
1
+
F
y
2
 W
g
= 0
=
F
1
sin
θ
1
+
F
2
sin
θ
2
 W
g
= 0 (2)
Solution:
Using Eqs. 1 and 2, we have
F
x
2
=
F
1
cos
θ
1
(1)
= (440 N) cos 36
◦
= 355
.
967 N
,
and
F
y
2
=
F
3

F
1
sin
θ
1
(2)
= 634 N

(440 N) sin 36
◦
= 634 N

258
.
626 N
= 375
.
374 N
,
so
F
2
=
radicalBig
(
F
x
2
)
2
+ (
F
y
2
)
2
=
radicalBig
(355
.
967 N)
2
+ (375
.
374 N)
2
= 517
.
319 N
.
Question 4, chap 5, sect 5.
part 2 of 2
10 points
b) What is the angle
θ
2
which the lefthand
cable makes with respect to the ceiling?
Correct answer: 46
.
5201
◦
(tolerance
±
1 %).
Explanation:
Using Eq. 2, we have
θ
2
= arctan
parenleftbigg
F
y
1
F
x
1
parenrightbigg
= arctan
parenleftbigg
258
.
626 N
355
.
967 N
parenrightbigg
= 46
.
5201
◦
.
Question 5, chap 5, sect 6.
part 1 of 3
10 points
A block of mass 1
.
84572 kg lies on a friction
less table, pulled by another mass 3
.
62536 kg
under the influence of Earth’s gravity.
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 Spring '08
 Turner
 Physics, Force, Friction, Work, Correct Answer, Fontenot, BRIAN

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