§
5 Problem 24 (
a
): Let up be the positive direction. A student with weight
w
= 550 N has mass
m
=
w
g
= 56 kg.
The forces on the student are the weight
w
down and a normal force
N
= 450 N up. (A bathroom scale
doesn’t actually measure your weight: it measures the force with which it is squeezed between your feet and
the floor, which is the normal force between you and the floor. This is usually your weight.) Thus Newton’s
second law says that
N

w
=
ma
⇒
a
=
N

w
m
=
(450 N)

(550 N)
56 kg
=

1
.
8 m
/
s
2
.
The elevator is accelerating down.
(
b
): Now we have
a
=
N

w
m
=
(670 N)

(550 N)
56 kg
= 2
.
1 m
/
s
2
.
The elevator is accelerating up.
(
c
): Now we have
a
=
N

w
m
=
(0 N)

(550 N)
56 kg
=

9
.
8 m
/
s
2
.
The elevator is in free fall, and the student should worry.
(
d
): For the elevator and student combined, there are two forces: the tension
T
of the cable pulling up and the
weight
W
=
Mg
= (850 kg)(9
.
8 m
/
s
2
) = 8300 N down. Newton’s second law says that
T

Mg
=
Ma
⇒
T
=
M
(
g
+
a
)
.
For part (
a
) we have
a
=

1
.
8 m
/
s
2
, so
T
= (850 kg)(8
.
0 m
/
s
2
) = 6800 N
.
For part (
c
) we have
a
=

9
.
8 m
/
s
2
, so
T
= 0.
§
5 Problem 33 (
a
):
Let us use co¨ordinates parallel and perpendicular to the slope.
Let up be the positive
directions. The angle of the slope is
θ
= Tan

1
parenleftbigg
2
.
50 m
4
.
75 m
parenrightbigg
= 27
.
8
◦
.
Let
m
1
be the larger, bottom mass and
m
2
be the smaller, upper mass. The perpendicular forces on
m
2
are
N
2

m
2
g
cos
θ
= 0
⇒
N
2
=
m
2
g
cos
θ,
where
N
2
is the normal force between
m
1
and
m
2
.
Here we have used the fact that the perpendicular
acceleration is 0. The parallel acceleration is also 0, because the problem states that the boxes are moving
with constant velocity. The parallel forces are
F
f,
2

m
2
g
sin
θ
= 0
⇒
F
f
=
m
2
g
sin
θ.
Here
F
f,
2
is the force of friction exerted on
m
2
by
m
1
, and we guess that it points uphill. (If we had guessed
wrong, we would just have gotten a negative number.)
Now we examine the perpendicular forces on
m
1
:
N
1

N
2

m
1
g
cos
θ
= 0
⇒
N
1
=
N
2
+
m
1
g
cos
θ
= (
m
1
+
m
2
)
g
cos
θ,
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
where
N
1
is the normal force between
m
1
and the ramp.
Note that because
m
1
is pushing
m
2
up with
force
N
2
,
m
2
is pushing
m
1
down with the same strength. The parallel forces are
F
f,
1

F
f,
2
+
F

m
1
g
sin
θ
= 0
,
where
F
is the force with which we are pulling, and
F
f,
1
is the friction force. Note that because the boxes
are sliding down the slope the friction points up the slope.
Again, as
m
1
pushes
m
2
with force
F
f,
2
,
m
2
pushes
m
1
an equal and opposite amount. For kinetic friction we know that
F
f,
1
=
μ
k
N
1
=
μ
k
(
m
1
+
m
2
)
g
cos
θ,
so that the force we need to apply is
F
=
m
1
g
sin
θ
+
F
f,
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '08
 Chamberlin
 Force, Friction, Mass, Work, Sin, Cos

Click to edit the document details