The Essential Galileo (54 page)

[154] If what precedes is clear, you will not hesitate, I think, to admit that the two prisms
AD
and
DB
are in equilibrium about the point
C
since one-half of the whole body
AB
lies on the right of the suspension
C
and the other half on the left; in other words, this arrangement is equivalent to two equal weights disposed at equal distances. I do not see how anyone can doubt, if the two prisms
AD
and
DB
were transformed into cubes, spheres, or any other figure whatever, and if
G
and
F
were retained as points of suspension, that they would remain in equilibrium about the point
C;
for it is only too evident that change of figure does not produce change of weight so long as the quantity of matter does not vary. From this we may derive the general conclusion that any two heavy bodies are in equilibrium at distances that are inversely proportional to their weights.

This principle established, I desire, before going further, to call your attention to the fact that these forces, resistances, moments, figures, etc., may be considered either in the abstract, dissociated from matter, or in the concrete, associated with matter. Hence the properties which belong to figures that are merely geometrical and nonmaterial must be modified when we fill these figures with matter and so give them weight. Take, for example, the lever
BA
, which, resting upon the support
E
, is used to lift a heavy stone
D
. The principle just demonstrated makes it clear that a force applied at the extremity
B
will just suffice to balance the resistance offered by the heavy at
D
the same ratio as
body
D
provided this force bears to the force at the distance
AC
bears to the distance
CB;
and this is true so long as we consider only the moments of the single force at
B
and of the resistance at
D
, treating the lever as an immaterial body devoid of weight. But if we take into account the weight of the lever itself—an instrument that may be made either of wood or of iron—it is manifest that, when this weight has been added to the force at
B
, [155] the ratio will be changed and must therefore be expressed in different terms. Hence before going further let us agree to distinguish between these two points of view: when we consider an instrument in the abstract, i.e., apart from the weight of its own matter, we shall speak of
taking it in an absolute sense;
but if we fill one of these simple and absolute figures with matter and thus give it weight, we shall refer to such a material figure as a
moment, or compound force
.

S
AGR.
    I must break my resolution about not leading you off into a digression, for I cannot concentrate my attention upon what is to follow until a certain doubt is removed from my mind. That is, you seem to compare the force at
B
with the total weight of the stone
D
, a part of which—possibly the greater part—rests upon the horizontal plane, so that …
¹²

S
ALV.
    I understand perfectly; you need go no further. However, please observe that I have not mentioned the total weight of the stone. I spoke only of its force at the point
A
, the extremity of the lever
BA;
this force is always less than the total weight of the stone and varies with its shape and elevation.

S
AGR.
    Good; but there occurs to me another question about which I am curious. For a complete understanding of this matter, I should like you to show me, if possible, how one can determine what part of the total weight is supported by the underlying plane and what part by the end
A
of the lever.

S
ALV.
    The explanation will not delay us long, and I shall therefore be glad to grant your request. In the following figure, let us understand that the weight having its center of gravity at
A
rests with the end
B
upon the horizontal plane and with the other end upon the lever
CG.
Let
N
be the fulcrum of the lever to which a force is applied at
G.
Drop the perpendiculars,
AO
and
CF
, from the center
A
and the end
C.
Then, I say, the moment of the entire weight bears to the moment of the force at
G
a ratio compounded of the ratio between the two distances
GN
and
NC

and the ratio between
FB
and
BO
.

Lay off a distance
X
such that the ratio of
FB
to
BO
is the same as that of
NC
to
X
. But since the total weight
A
is supported by the two forces at
B
and at
C
, [156] it follows that the force at
B
is to that at
C
as the distance
FO
is to the distance
BO
. Hence, by addition, the sum of the forces at
B
and
C
, that is, the total weight
A
, is to the force at
C
as the line
FB
is to the line
BO
, that is, as
NC
is to
X
. But the force applied at
C
is to the force applied at
G
as the distance
GN
is to the distance
NC.
Hence it follows, by perturbed equidistance of ratios,
¹³
that the entire weight
A
is to the force applied at
G
as the distance
GN
is to
X
. But the ratio of
GN
to
X
is compounded of the ratio of
GN
to
NC
and of
NC
to
X
, that is, of
FB
to
BO.
Hence the weight
A
bears to the supporting force at
G
a ratio compounded of that of
GN
to
NC
and of
FB
to
BO
. This is what had to be demonstrated.

Let us now return to our original subject. If what has hitherto been said is clear, it will be easily understood why the following (
Proposition 1
) is true:
A prism or solid cylinder of glass, steel, wood, or other breakable material, which is capable of sustaining a very heavy weight when applied longitudinally, is (as previously remarked) easily broken by the transverse application of a weight that may be much smaller in proportion as the length of the cylinder exceeds its thickness.

Let us imagine a solid prism
ABCD
fastened into a wall at the end
AB
, and supporting a weight
E
at the other end; understand also that the wall is vertical and that the prism or cylinder is fastened at right angles to the wall. It is clear that if the prism breaks, fracture will occur at the point
B
where the edge of the slot in the wall acts as a fulcrum. The length
BC
acts as the part of the lever to which the force is applied. The thickness
BA
of the solid is the other arm of the lever in which is located the resistance. This resistance opposes the separation of the part of the solid
BD
lying outside the wall from the portion lying inside. From the preceding, it follows that the moment of the force applied at
C
bears to the moment of the resistance found in the thickness of the prism (i.e., in the attachment of the base
BA
to its contiguous parts) the same ratio which the length
CB
bears to half of
BA.
[157] Now if we call absolute resistance to fracture that offered to a longitudinal pull (in which case the stretching force moves by the same amount as the stretched body), then we can say that the absolute resistance of the prism
BD
is to the breaking load placed at the end of the lever
BC
in the same ratio as the length
BC
is to the half of
AB
in the case of a prism, or the radius in the case of a cylinder. This is our first proposition.

Note that in what has here been said the weight of the solid
BD
itself has been left out of consideration, or rather, the prism has been assumed to be devoid of weight. But if the weight of the prism is to be taken into account in conjunction with the weight
E
, we must add to the weight
E
one half that of the prism
BD.
Thus, for example, if the latter weighs two pounds and the weight
E
is ten pounds, we must treat the weight
E
as if it were eleven pounds.

S
IMP.
    Why not twelve?

S
ALV.
    The weight
E
, my dear Simplicio, hanging at the extreme end
C
acts upon the lever
BC
with its full moment of ten pounds. If suspended at the same point, the solid
BD
would also exert its full moment of two pounds. But as you know, this solid is uniformly distributed throughout its entire length,
BC
, so that the parts which lie near the end
B
are less effective than those more remote. Accordingly, if we strike a balance between the two, the weight of the entire prism may be considered as concentrated at its center of gravity, which lies at the midpoint of the lever
BC.
But a weight hung at the extremity
C
exerts a moment twice as great as it would if suspended from the middle. Therefore, [158] if we consider the moments of both as located at the end
C
, we must add to the weight
E
one-half that of the prism.

S
IMP.
    I understand perfectly. Moreover, if I am not mistaken, the force of the two weights
BD
and
E
, thus disposed, would exert the same moment as would the entire weight
BD
together with twice the weight
E
suspended at the middle of the lever
BC.

S
ALV.
    Precisely so, and a fact worth remembering. Now we can readily understand
Proposition 2
:
How and in what proportion a rod, or rather a prism, whose width is greater than its thickness offers more resistance to fracture when the force is applied in the direction of its width than in the direction of its thickness.

For the sake of clarity, take a ruler
ad
whose width is
ac
and whose thickness,
cb
, is much less than its width. The question now is why will the ruler, if stood on edge, as in the first figure, withstand a great weight
T
, while, when laid flat, as in the second figure, it will not support the weight
X
, which is less than
T.
The answer is evident when we remember that in the one case the fulcrum is at the line
bc
, and in the other case at
ca
, while the distance at which the force is applied is the same in both cases, namely, the length
bd
. But in the first case the distance of the resistance from the fulcrum—half the line
ca
—is greater than in the other case where it is only half of
bc
. Therefore, the weight
T
is greater than
X
in the same ratio as half the width
ca
is greater than half the thickness
bc
, since the former acts as a lever arm for
ca
, and the latter for
cb
, against the same resistance, namely, the strength of all the fibers in the cross section
ab
. We conclude, therefore, that any given ruler, or prism, whose width exceeds its thickness, will offer greater resistance to fracture when standing on edge than when lying flat, and this in the ratio of the width to the thickness.