Page images



70. To determine the horizontal thrust of the terrace, whose vertical section is BCHK, against the wall whose section is ABCD, and the momentum of the thrust to overturn the wall about the angle A.

[blocks in formation]

If it be required to support a terrace by a vertical wall, it must be constructed so as to counteract the horizontal thrust of the prismatic mass of earth which lies above the surface of a bank that would be itself supported. But this prismatic mass is partly supported by friction, and we must therefore ascertain how much of the horizontal thrust is counteracted by friction.

Suppose a weight W to be placed on a plane, inclined to the vertical at an angle i; and let H be the horizontal force, which, with the friction, just sustains the weight W. Resolve each of the forces W, H into two others, the one parallel and the other perpendicular to the plane; and those parallel to the plane act in opposite directions, while those perpendicular to the plane concur in direction; hence we have


[ocr errors]


force parallel to the plane

= W cos i H sin i
W sin i + H cos i.

force perpendicular to the plane

And the first of these forces must be precisely equal to the friction; that is, equal to a force that will just support the weight upon the plane; hence W cos i H sin if W sin i + f H cos i .. H = cos i -f sin i W = sin i+fcos i If then the weight W were sustained the weight W against the wall would be

1 - ftan i w


by a wall, the horizontal thrust of
1 - ftan i
tan i +f



Now to apply this to the investigation of the horizontal thrust of the prism BCH, we shall put BC= a, a variable part Cb=x, bb' = dx, and s the specific gravity of the earth. Then the area of bb'hh': xdx tan i, and its weight = sxdx tan i; hence the horizontal thrust against bb' will be 1 - ftan i sxdx tan i - ftan i tan i +f 1+fcot i where M = - ftan i ; hence, integrating, we have 1+fcot i fos Mxdx=a's M = whole horizontal thrust of triangle BCH.




Again, the length of the lever Bb = a x, and the moment* of the thrust of the element bb'hh' = sMx (a — x) dx = as Mxdx

sMx2dx; hence


* If lines be expressed numerically, the product of a force acting on a lever, and the perpendicu

lar from the axis of motion on its direction is called the moment of that force.

fas Mxdx-f°sMx2dx = 4a3s M — a3s M horizonal thrust.

The expression


-ftan i 1 + f cot i between these limits there is a value which gives both the horizontal thrust and its moment a maximum. Let then



u =



tan. is


Hence M

2 tan i


1 - ftan i
1 + f cot i

.. horizontal thrust =
and moment of thrust =

and tan

[ocr errors]

1-ƒtan i
= maximum, and differentiating we have
1 + f cot i

f sec2 i (1+f cot i) + ƒ cosec2 i (1 —ƒ tan i)
(1 + f cot i)2

sec2 i (1+fcot i) = cosec2 i (1 — ftan i)
tan i
=−ƒ+ √i + ƒ2




71. The angle whose tangent is -ƒ + √1 + ƒ2 is just half of that whose



1 2

or tan-1(−ƒ + √1 +ƒ3) =

For since tan 2i =

2(f+I+ƒo) 2(−ƒ+ √ + ƒ2)
1 − (−ƒ+ √1 + ƒ22)2 2ƒ(−ƒ+ √1 +ƒ3)


12 isn

is the angle of the slope which the earth would naturally assume







will vanish when tan i : 0, or tan i =



9 hence x = 40

if unsustained by any wall.

For if i be the inclination of a plane to the vertical, and g the accelerating force of gravity; then the force g resolved into two, parallel and perpendicular to the plane, gives g cos i and g sin i; hence the friction =ƒg sin i, and being counteracted by the force g cos i, we must have

= fa3s M = moment of the whol

sec2 i


cosecitan2 = (−ƒ + √T + ƒ2)2
4a2s( −ƒ + √1 + ƒ2)2 = 4a2s tan2 i.
fa3s( −ƒ + √1 +ƒ2)2 = ¿a3s tan2 i.

[ocr errors][merged small][merged small]

Hence if BK be the natural slope of loose earth, and BH bisect the angle KBC; then the prismatic mass CBH will exert the greatest force against the vertical wall BC.

72. In loose earth the natural slope is about 60° from the vertical, and in tenacious earth this angle is about 54°; hence in the former case i = 30; tan i = tan 30° = √1⁄2=√/3, and in the latter i= 27°, tan i = tan 27°√, nearly. Therefore, for loose earth, the horizontal thrust ja2s, and its moment a3s, and for tenacious earth, the horizontal thrust is 4a3s, and its momenta3s. Now put AB the breadth of the wall = x, BC= a, and the specific gravity = S; then the moment of the resistance of the wall is = ax2S, which, in the case of equilibrium, must be equal to the moment of the horizontal thrust; hence, for tenacious earth we have


ax2S= a3s

1 24

[ocr errors]


[ocr errors]

Ex. 1. Let S = 2520, and s = 1600; then we have






[ocr errors]


or f = cot i.

[ocr errors]



= 23= nearly.

9 40

of the height of the rectangular wall;


Ex. 2. Let the wall be triangular, as in the annexed figure, and let x = its breadth; then the moment of the resistance will be hence we must have

× ‡ ax$ = fax2S;

[merged small][ocr errors][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small]
[ocr errors]

A is =

[ocr errors]


[ocr errors]
[ocr errors]

LET ABCD be half the arch, and DEFG the pier. From the centre of gravity K of the arch draw KL perpendicular to the horizon. Then the weight of the arch in direction KL will be to the horizontal push at A, in direction LA, as KL to LA.

For the weight of the arch in direction KL, the horizontal push or lateral pressure in direction LA, and the push in direction KA, will be as the three sides



[ocr errors]
[ocr errors]



73. To determine the thickness of a pier necessary to support a given arch



of the height of the triangular wall.


= 282 nearly =


[ocr errors]


KL, LA, KA. So that, if A denote the weight or area of the arch;


will be its force at A in the direction LA; and
the lever GA to overset the pier, or to turn it about the point F.

Again, the weight or area of the pier, is as EF. FG; and therefore EF. FG. FG, or EF. FG2, is its effect on the lever FG, to prevent the pier from being overset; supposing the length of the pier, from point to point, to be no more than the thickness of the arch.

But that the pier and arch be in equilibrio, these two effects must be equal.

Therefore we have EF. FG 2 = 2


of the pier is FG = √ˆ ̃EF. KL × A.

[merged small][ocr errors]



[ocr errors]



GA. A its effect on


[ocr errors]

Example 1. Suppose the arc ABN to be a semicircle; and that DC or AO = 45, BC= 6, and GA≈ 18 feet. Then KL will be found 40, AL = 15 also, the area ABCD or A = 7043. Therefore FG = 36.15

nearly, and EF = 69;


704 11 nearly, which is the thickness


of the pier.


GA. A, and consequently the thickness

Example 2. Suppose, in the segment ABN, AN = 100, OB = 411, BC= 61⁄2, and AG 10. Then EF = 58, KL = 35, AL = 15 nearly, and ABCD or /2GA. AL A 842. Therefore FG = A = 842 = 11 EF.KL nearly, the thickness of the pier in this case.





1. A body is said to be in motion when it is continually changing its position in space.

2. Motion is said to be uniform when the spaces described in equal successive intervals of time are equal, and variable when these spaces are unequal.

3. The velocity of a body is the space it would describe in a unit of time, were the motion to become uniform at the commencement of that unit.

4. Motion is said to be accelerated when the velocity continually increases, and retarded when it continually decreases; and an accelerating or retarding force is said to be uniform or variable, according as the increments or decrements of velocity in equal times are equal or unequal.

5. The momentum or quantity of motion of a body is the sum of the motions of all its particles; and, as the motion of a particle is measured by its velocity, and the number of particles in a body constitutes its mass; hence the momentum will be equal to the product of the mass and velocity, when all the particles move with the same velocity.

6. Inertia is the opposition offered by a body to a change of state, either of rest or of motion, by the action of a force impressed upon it.

7. If a system of particles, m, m1, m,, ... revolve round an axis, and r, r', r",...... be their respective distances from that axis; then mr2 + m1r12 + m2r1/2 + . or (mr) is called the moment of inertia of the system.



8. If a spherical body strike or act obliquely on a plane surface, the force or energy of the stroke or action, is as the sine of the angle of incidence.

Or, the force on the surface is to the same if it had acted perpendicularly, as the sine of incidence is to radius.

Let AB express the direction and the absolute quantity of the oblique force on the plane DE; or let a given body A, moving with a certain velocity, impinge on the plane at B; then its force will be to the action on the plane, as radius to the sine of the angle ABD, or as AB to BC, drawing BC perpendicular, and AC parallel to DE.


For, by Prop. II., the force AB is equivalent to the two forces AC, CB; of which the former AC does not act on the plane, because it is parallel to it. The plane is therefore only acted on by the direct force CB, which is to AB as the sine of the angle BAC, or ABD, to radius.

Corollary. 1. If a body act on another, in any direction, and by any kind of force, the action of that force on the second body, is made only in a direction perpendicular to the surface on which it acts.

For, the force in AB acts on DE only by the force CB, and in that direction. Corollary. 2. If the plane DE be not absolutely fixed, it will move after the stroke, in the direction perpendicular to its surface. For it is in that direction that the force is exerted.


9. If one body A strike another body B, which is either at rest or moving towards the body A, or moving from it, but with a less velocity than that of A; then the momenta, or quantities of motion of the two bodies, estimated in any one direction, will be the very same after the stroke that they were before it.

For, because action and re-action are always equal, and in contrary directions, whatever momentum the one body gains one way by the stroke, the other must just lose as much in that same direction; and therefore the quantity of motion in that direction, resulting from the motions of both the bodies, remains still the same as it was before the stroke.

Thus, if A with a momentum of 10, strike B at rest, and communicate to it a momentum of 4, in the direction AB. Then A will have only

a momentum of 6 in that direction; which, together with the momentum of B, viz. 4, make up still the same momentum between them as before, namely, 10. If B were in motion before the stroke, with a momentum of 5, in the same direction, and receive from A an additional momentum of 2; then the motion of A after the stroke will be 8, and that of B, 7, which between them make 15, the same as 10 and 5, the motions before the stroke.


Lastly, if the bodies move in opposite directions, and meet one another, namely, A with a motion of 10, and B, of 5; and A communicate to B a motion of 6 in the direction AB of its motion. Then, before the stroke, the whole motion from both, in the direction of AB, is 10-5 or 5; but after the stroke, the motion of A is 4 in the direction AB, and the motion of B is 6-5 or 1 in the same direction AB; therefore, the sum 4+ 1, or 5, is still the same motion from both as it was before.


10. The motion of bodies included in a given space, is the same, with regard to each other, whether that space be at rest, or move uniformly in a right line.

For, if any force be equally impressed both on the body and the line in which it moves, this will cause no change in the motion of the body along the right line. For the same reason, the motions of all the other bodies, in their several directions, will still remain the same. Consequently, their motions among themselves will continue the same, whether the including space be at rest, or be moved uniformly forward; and therefore, their mutual actions on one another must also remain the same in both cases,

« PreviousContinue »