Environmental Engineering Reference
In-Depth Information
which proves that the centre-of-mass lies on a line joining the two
masses.
(b) The position of the centre-of-mass is given by:
1
3
(
a
=
+
+
R
b
c
).
Now the median from
a
that intersects the side between
b
and
c
has the
vector equation
λ
1
a
λ
2
(
b
r
=
a
+
2
(
b
+
c
)
−
=
a
+
+
c
−
2
a
) .
We can write
1
3
(
b
R
=
a
+
+
c
−
2
a
)
showing that
R
lies on this median. Repeat for the other two medians.
2.2 The total mass of the system is 80 g so
13
8
j
m
.
1
80
(
130
i
25
8
R
=
+
250
j
)
=
i
+
It doesn't matter to which particle the force is applied, the resulting acceler-
ation of the centre-of-mass is the same
3
0
.
08
R
37
.
5
i
ms
2
.
=
i
=
=
2.3 Calculate the rate of change of momentum to obtain
F
20 N.
2.4 The friction between the prisoner's hands and the rope acts as a brake pro-
viding an upwards force on the prisoner. For the rope to remain in static
equilibrium the tension must balance the frictional force. The maximum ten-
sion is
T
max
=
600 N. So the acceleration
a
is obtained from
ma
=
mg
−
T
max
.
Once you have the acceleration it is a simple problem to find the speed at
impact:
2
h
g
T
max
m
35 km hr
−
1
9
.
6m s
−
1
.
v
=
−
=
=
2.5 Draw a free-body diagram and resolve components of the forces parallel and
perpendicular to the slope. The force applied by the Egyptians parallel to the
slope is
mg
sin 20
◦
=
F
=
ma
+
18
.
3kN
.
