Problem 1.
(a) Assume a buoyant force for the object to be weighed to be FB and
the buoyant force for the counter weight to be FB0 . Balancing forces on the
equal arm balance results in
FB = Fg0
Fg
FB0 ;
where Fg = mg and Fg0 = m0 g are the actual weights (in vaco) for the object
and counter weight respectively. The buoyant forces are
FB =
air V
g and FB0 =
air V
0
g;
where V and V 0 are the volumes of the object and counter weight respectively.
Substituting these results into the equilibrium force equation results in
Fg = Fg0 + FB
FB0 = Fg0 + (V
0
The density of the counter weight is
Fg = Fg0 + (V
m0 = 0 )
V 0)
air g:
= m0 =V 0 : Substituting for V 0 yields
0
air g = Fg + V
Fg0
0g
air g:
This is the desired result.
(b) If the density of the object is equal to that of the counter weight then
0
m =V 0 = m=V = and
Fg = Fg0 + (m=
m0 = 0 )
air g:
= Fg0 + (m
Fg = Fg0 + Fg
Fg0
air
Fg0
air
= 0 ! Fg = Fg0 :
Fg
1
m0 ) g
air
This is the desired result.
Problem 2 - Chapter 19 problem 74.
Consider a cone of length ` and faces of radii R1 and R2 which are held
at constant temperatures T1 and T2 respectively (see …gure 19-27). Now a
more general expression for the rate of heat ‡ow is
H=
kA
1
dT
:
dx
Assuming that the sides of the cone are perfectly insulated then the heat
‡ow is only along the length of the cone. De…ning the origin to be x = 0 at
the face with radius R1 ; the radius R as a function of x is given by
R (x) =
R2
R1
`
x + R1 :
Hence the cross sectional area as a function of x is
R2
A (x) =
R1
`
2
x + R1
The equation for the heat ‡ow can now be written as
H=
k
R2
R1
`
2
x + R1
dT
:
dx
Noting that H is constant (otherwise there would have to be a source of heat
inside the cone) we …nd
kdT =
((R2
H
dx:
R1 ) x=` + R1 )2
Integrating this expression from x = 0 (where T = T1 ) to x = ` (where T = T2 )
we …nd
Z T2
Z `
dx
k
dT = H
R1 ) x=` + R1 )2
T1
0 ((R2
k (T2
k (T2
k (T2
H`2
T1 ) =
(R2 R1 )2
H`
T1 ) =
(R2 R1 )
H`
T1 ) =
(R2 R1 )
`
1
x + `R1 = (R2 R1 ) 0
1
1
R2 R1
R1 R2
H`
=
:
R2 R1
R2 R1
Solving for H we …nd
H = kR1 R2 (T1
T2 ) =`:
Note that if T1 > T2 then the heat ‡ow is in the positive direction, while if
T1 < T2 then the reverse is true.
2