Physics 112 - HW #5 Solutions
5-42
When ball's speed v = v
t
/2 =
mg/D /2 (half its terminal speed), f
d
= Dv
2
= mg/4.
(a)
Going up:
mg
f
d
m
Σ
F
= m
a
(
↓
+
):
mg + f
d
= ma
=
mg +
mg
4
=
5mg
4
⇒
a
=
5
4
g
↓
(b)
Going down:
mg
f
d
m
Σ
F
= m
a
(
↓
+
):
mg - f
d
= ma
=
mg -
mg
4
=
3mg
4
⇒
a
=
3
4
g
↓
5-47
The free-body diagram shows the forces acting on the plane
while turning, as viewed from the front or rear.
The plane's
acceleration is
a = v
2
/R
in the +x direction (
←
).
v
=
240 km/h
10
3
m
1 km
1 h
3600 s
=
66.7 m/s
Write Newton's 2nd Law,
Σ
F
= m
a
:
+x:
F
lift
sin
θ
= ma
x
= m
v
2
R
(1)
+y:
F
lift
cos
θ
- mg
=
ma
y
=
0
⇒
F
lift
cos
θ
=
mg
(2)
Divide equation (1) by equation (2) to get:
F
lift
sin
θ
F
lift
cos
θ
=
tan
θ
=
v
2
Rg
=
(66.7 m/s)
2
(1200 m)(9.8 m/s
2
)
=
0.378
⇒
θ
=
21°
5-75
(a)
Draw a free-body diagram for all forces acting on the crate, including the friction
force f.
We don't yet know whether
f
is static or kinetic.
Write out
Σ
F
= m
a
:
y:
N - mg
=
ma
y
= 0
⇒
N = mg
x:
f
=
ma
x
= ma
If
f
is static friction, then
f = ma = f
s
≤
µ
s
N
=
µ
s
mg ,
so that . . .
This
preview
has intentionally blurred sections.
Sign up to view the full version.

This is the end of the preview.
Sign up
to
access the rest of the document.
- Fall '07
- LECLAIR,A
- Physics, mechanics, Force, Friction, free-body diagram, Write Newton
-
Click to edit the document details