Environmental Engineering Reference
In-Depth Information
The normal force must balance the component of the block's weight normal
to the surface, i.e.
mg
cos 20
◦
=
N
=
46
.
0kN
.
2.6 Newton's Second Law, and the expression for centripetal acceleration, gives
the normal force on the skier to be
m
g
v
2
r
N
=
−
=
214 N
.
This is directed upwards, but must be equal in magnitude, and opposite in
direction to the force that the skier exerts on the snow, which is 214 N
downwards.
2.7 The acceleration of both blocks is given by
F
m
1
+
a
=
m
2
.
The second block accelerates due to the normal reaction force between the
blocks. This has the value
m
2
F
m
1
+
3
4
N
.
F
=
m
2
=
2.8
sfasfd
(a) The only horizontal force on the upper block (
m
1
) is friction. This has
a maximum value given by
µ
s
m
1
g
resulting in an acceleration
µ
s
g
.In
this case both blocks have the same acceleration and the force on the
whole system is 17.7 N.
(b) Both blocks have the same acceleration,
a
1
.
47 ms
−
2
.
Since the upper
block accelerates due to the force of friction we calculate the frictional
force to be 2 kg
=
1
.
47 ms
−
2
2
.
94 N.
(c) In this case the top blocks slips and we have kinetic friction that causes
the top block to accelerate. Thus the acceleration of the top block is
µ
k
g
×
=
1
.
96 ms
−
2
. The bottom block accelerates due to the resultant of
the applied force and the opposing friction due to the top block. Its
acceleration is therefore 7
.
87 ms
−
2
.
=
2.9 Obtain an equation for the net force
F
on the system as a whole by con-
sidering the change in momentum of the “rocket plus exhaust” system. If
amass
m
is ejected you should find that the change in momentum is
p
(m)
u
,
from which the rocket equation follows after divid-
ing by
t
and taking the limit that
t
goes to zero. Note that there is a sign
change because
m
=
m
v
+
d
m
d
t
≈−
d
t,
i.e. the mass of the rocket is decreasing with
time. For
F
=
0
we can write
t
f
u
t
f
t
i
d
v
d
t
d
t
1
m
d
m
d
t
m
i
m
f
=
d
t
=−
u
ln
.
t
i
