6. In the preceding Example if additional support be given by means of a string fastened to the body and to a point in the plane, so that the string may be parallel to the plane, find at what inclination of the plane the string would break, supposing the string would break on a smooth inclined plane at an inclination of 45o.

7. If the height of a rough inclined plane be to the length as 3 is to 5, and a Weight of 15 lbs. be supported by friction alone, find the force of friction in lbs.

8. If the height of a rough inclined plane be to the length as 3 is to 5, and a weight of 10 lbs. can just be supported by friction alone, shew that it can just be drawn up the plane by a force of 12 lbs. along the plane.

5

9. Find the force along the plane required to draw a weight of 25 tons up a rough inclined plane, the coefficient of friction being and the inclination of the plane being such that 7 tons acting along the plane would support the Weight if the plane were smooth.

12'

10. Find the force in the preceding Example, supposing it to act at the most advantageous inclination to the plane.

11. A ladder inclined at an angle of 60o to the horizon rests between a rough pavement and the smooth wall of a house. Shew that if the ladder begin to slide when a man has ascended so that his centre of gravity is half way up, then the coefficient of friction between the foot of the ladder and the pavement is√3.

1

6

12. A heavy beam rests with one end on the ground, and the other in contact with a vertical wall. Having given the coefficients of friction for the wall and the ground, and the distances of the centre of gravity of the beam from the ends, determine the limiting inclination of the beam to the horizon.

XX. Miscellaneous Propositions.

262. In the present Chapter we shall give some miscellaneous propositions of Statics.

263. We have hitherto confined ourselves almost entirely to the equilibrium of forces which act in a plane: the following Articles will contain some propositions explicitly relating to forces which are not all in one plane.

264. To find the resultant of three forces which act on a particle and are not all in one plane.

Let OA, OB, OC represent three forces in magnitude and direction which act on a particle at O. Let a parallelepiped be formed having these straight lines as edges; then the diagonal OD which passes through O will represent the resultant in magnitude and direction.

B

E

D

A

For OE, the diagonal passing through O of the parallelogram OAEB, represents in magnitude and direction the resultant of the forces represented by OA and OB. Also OEDC is a parallelogram, and OD, the diagonal passing through O, represents in magnitude and direction the resultant of the forces represented by OE and OC, that is, the resultant of the forces represented by OA, OB, and OC.

265. The preceding investigation is only a particular case of the general process given in Art. 52, but on account of its importance it deserves special notice. As we can thus compound three forces into one, so on the other hand we can resolve a single force into three others which act in assigned directions. Most frequently when we have thus to resolve a force the assigned directions are mutually

T. M.

12

at right angles; that is with the figure of Art. 264, the angles AOB, BOC, COA are right angles. The angle OČD is then a right angle, so that OČ=OD cos COD: thus when the three components are mutually at right angles a component is equal to the product of the resultant into the cosine of the angle between them.

Also by Euclid, I. 47, we have

OD2=OC2+ CD2= OC2 + OE2=OC2+OB2 + OA2 :

thus when the three components are mutually at right angles the square of the resultant is equal to the sum of the squares of the three components.

266. The process of Art. 52 for determining the resultant of any number of forces acting on a particle is applicable whether the forces are all in one plane or not; the process in Art. 56 assumes that the forces are all in one plane: we shall now extend the latter process to the case of forces which are not all in one plane.

267. Forces act on a particle in any directions: required to find the magnitude and direction of their resultant.

Let O denote the position of the particle; let Op, Oq, Or, Os,... denote the directions of the forces; let P, Q, R, S,... denote the magnitudes of the forces which act along these directions respectively.

Draw through O three straight lines mutually at right angles; denote them by Ox, Oy, Oz: and resolve each force into three components along these straight lines, by Art. 265. Thus P may be replaced by the following three components: P cos pox along Ox, P cos pOy along Oy, and P cos poz along Oz. Similarly Q may be replaced by Qcos qOx along Ox, Q cos qoy along Oy, and Qcos qoz along Oz. And so on.

Let X denote the algebraical sum of the components along Ox; so that

X=P cos pOx+ Q cos qOx + R cos rOx+S cos sOx+...

Similarly let Y and Z denote the algebraical sums of the components along Oy and Oz respectively.

Thus the given system of forces is equivalent to the three forces X, Y, Z which act along three straight lines mutually at right angles.

Let

denote the resultant of the given system of forces,

and Ov its direction; then, by Art. 265,

Thus the magnitude and the direction of the resultant are determined.

268. Since V cos v0x=X, the resolved part of the resultant in any direction is equal to the sum of the resolved parts of the components in that direction: see Arts. 44 and 90.

269. In Art. 39 we have given the conditions under which three forces acting on a particle will maintain it in equilibrium; we will now present these conditions in a slightly different form, and then demonstrate a corresponding result for the case of four forces which are not all in one plane.

270. OA, OB, OC are three straight lines of equal length in one plane, and they are not all on the same side of any straight line in the plane passing through 0; forces P, Q, R respectively act along these straight lines such that

P

area of OBC

Q

=

R

:

area of OCA area of OAB these forces will maintain a particle at O in equilibrium. For the area of a triangle is half the product of two sides into the sine of the included angle; hence each force is proportional to the sine of the angle between the directions of the other two: and the proposition follows immediately from Art. 39.

271. OÂ, OB, OC, OD are four straight lines of equal length, no three of them being in the same plane, and they are not all on the same side of any plane passing through 0; forces P, Q, R, S respectively act along these straight lines such that

P

=

Q

=

R

=

S

vol. OBCD vol. OCDA vol. ODAB vol. OABC

these forces will maintain a particle at O in equilibrium.

Let p denote the length of the perpendicular from O on the plane BCD. Resolve each force into three along directions mutually at right angles, one direction being perpendicular to the plane BCD. The sum of the components of Q, R, and S in the direction perpendicular to the plane BCD

Let h denote the length of the perpendicular from 4 on the plane BCD. The component of P perpendicular to the plane BCD is Ph. Now the direction of the com

ΟΑ

ponent of P is opposite to the direction of the sum of the components of Q, R, and S by reason of the condition that the straight lines OA, OB, OC, OD are not all on the same side of any plane through O. Moreover by reason of the given ratios we have

P

vol. of OBCD

P+Q+R+S sumof vols of OBCD,OCDA,ODAB,OABC

Thus the algebraical sum of the components perpendicular to the plane BCD vanishes.