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Physics 121.6 2007/2008
Assignment 4 - Solutions
1. Chapter 5, Problem 2. A force F
r
applied to an object of mass m
1
produces an
acceleration of 3.00 m/s
2
. The same force applied to a second object of mass m
2
produces an acceleration of 1.00 m/s
2
. (a) What is the value of the ratio m
1
/m
2
?
(b) If m
1
and m
2
are combined into one object, what is the acceleration under the
action of the force F
r
?
Solution:
With F
r
= F ,
1 1
a m F = with a
1
=3.00 m/s
2
…
and
2

Transcript

Physics 121.6 2007/2008
Assignment 4 - Solutions
1.
Chapter 5, Problem 2.
A force
F
r
applied to an object of mass
m
1
produces an acceleration of 3.00 m/s
2
. The same force applied to a second object of mass
m
2
produces an acceleration of 1.00 m/s
2
. (a) What is the value of the ratio
m
1
/
m
2
? (b) If
m
1
and
m
2
are combined into one object, what is the acceleration under the action of the force
F
r
?
Solution:
With
F
r
=
F
,
11
amF
=
with
a
1
= 3.00 m/s
2
…
and
22
amF
=
with
a
2
= 1.00 m/s
2
…
(a) From
11
aF m
=
and from
22
aF m
=
So: 31m/s00.3
m/s00.1
22122121
====
aaaF aF mm
(b) What is
a
when ?
ammF
)(
21
+=
Using
and the result from (a):
m
2
= 3
m
1
22411411112111
m/s750.0)m/s00.3(
4)3()(
===⇒
=+=+=⇒
aaamammammam
- 1 -
Assignment 4 - Solutions
2.
Chapter 5, Problem 6.
A woman weighs 120 lb. Determine (a) her weight in Newtons and (b) her mass in kilograms.
Solution:
(a) Using 1 lb = 4.448 N Her weight is: N534lb1 N4.448lb120
=⎟ ⎠ ⎞⎜⎝ ⎛ =
W
(b) Assuming she is on the Earth where
g
= 9.80 m/s
2
,
W
=
mg
kg5.54
m/s80.9
N534
2
===⇒
gW m
3.
Chapter 5, Problem 14.
Three forces acting on an object are given by Nˆ00.2ˆ00.2
1
jiF
+−=
r
,
( )
Nˆ00.3ˆ00.5
2
jiF
−=
r
, and
N)ˆ0.45(
3
iF
−=
r
. The object experiences an acceleration of magnitude 3.75 m/s
2
. (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after 10.0 s? (d) What are the velocity components of the object after 10.0 s?
Solution:
(a)
Newton’s 2
nd
Law:
∑
=++=
aFFFF
rrrrr
m
321
So direction of
a
is in direction of net force.
r
( ) ( ) ( )
Nˆ)00.1(
ˆ)0.42(
ˆ0.45ˆ00.3ˆ00.5ˆ00.2ˆ00.2
jii ji jiF
−+−=
−+−++−=
∑
r
So direction of acceleration is given by
θ
, the angle counter clockwise around from the +
x
axis. Then:
°=⇒−−==
4.1810.4200.1tan
θ θ
x y
F F
(3
rd
quadrant) = 181
°
to 3 sig. figs. (b)
mam
==
∑
aF
rr
N0.42 N)00.1( N)0.42(
2222
=−+−=+=
∑
y x
F F
F
r
So kg2.11
m/s3.75 N0.42
2
===
∑
am
F
r
(c)
t t
i f
aavv
rrrr
=+=
since we know
0
=
i
v
r
. Therefore
f
v
r
is in same direction as
a
r
which is in same direction as
∑
F
r
. So speed = m/s5.37)s0.10)(m/s75.3(
2
====
at v
f f
v
r
(d)
m/s37.5)4.181cos()m/s5.37(cos
−=°==
θ
f xf
vv
m/s0.893)4.181sin()m/s5.37(sin
−=°==
θ
f yf
vv
- 2 -
Assignment 4 - Solutions
4.
Imagine you are standing on a cardboard box that is only just strong enough to support your weight. What would happen to it if you tried to jump into the air? The cardboard box would (A)
collapse. (B)
be unaffected. (C)
spring up as well. (D)
move sideways. (E)
None of these.
Solution:
When you try to jump into the air, you have to accelerate yourself vertically. To do this your legs act like springs which act like the interaction between your body and the box. Since there is an upward force on your body,
F
, giving you the upward acceleration, there must be an equal an opposite downward force on the box.
mg F a
+
y
Considering an FBD of the forces on the body:
mamgF mamgF F
y
+=⇒=−=
∑
Thus
F
is greater than your weight. Therefore the force downward on the box is greater than your weight, and is therefore greater than the box can support. Therefore the box will collapse. Answer
A
.
- 3 -
Assignment 4 - Solutions
5.
Chapter 5, Problem
20.
A bag of cement of weight 325 N hangs from three wires as shown in Figure P5.20. Two of the wires make angles
θ
1
= 60.0
°
and
θ
2
= 25.0
°
with the horizontal. Assuming the system is in equilibrium, find the tensions
T
1
,
T
2
, and
T
3
in the wires
.
Solution:
Since the bag of cement is in equilibrium: N3250
33
==⇒=−=
∑
W T W T F
y
W
θ
1
θ
2
T
1
T
2
T
3
A
x y
θ
1
θ
2
Since the join at point A is in equilibrium:
∑
=−=
0coscos
1122
θ θ
T T F
x
…
∑
=−+=
0sinsin
32211
T T T F
y
θ θ
…
From
2112
coscos
θ θ
T T
=
and inserting into
[ ][ ]
2113132111
3221111
tancossintancossinsincoscossin
θ θ θ θ θ θ θ θ θ θ
+=⇒=+⇒=+
T T T T T T T
[ ]
N296))tan(25.0cos(60.0)sin(60.0
N325
1
=°°+°
=⇒
T
Then N163)0.25cos(
)0.60cos(
296coscos
2112
=°°==
θ θ
T T
So
T
1
= 296 N,
T
2
= 163 N,
T
3
= 325 N.
- 4 -

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