286. A heavy rod rests with its ends on two given smooth inclined planes: required the position of equilibrium.

Let AB be the rod, AOM and BON the inclined planes; MON being a horizontal line.

The forces acting on the rod are the resistance of the plane at A, at right angles to OA, the resistance of the plane at B, at right angles to OB, and the weight vertically downwards through the centre of gravity of the rod.

Let

AC and BC be the lines of action of the resistances of the plane, and G the centre of gravity of the rod; then the directions of the three forces must meet at a point by Art. 41, so that G must be vertically under C. Join CG.

Let AG=a, BG=b, the angle AOM=a, and the angle BON=ẞ; and let O be the inclination of AB to the horizon, that is, the angle between AB and MN produced.

Since CA and CG are perpendicular to OA and OM respectively, the angle GCA=the angle AOM=a, Similarly the angle GCB=ß.

The angle OAB=a-0, and the angle ABO=ẞ+0, by Euclid, 1. 32.

287. As another example we will explain the balance called Roberval's Balance.

AC and BD are equal beams connected with the former beams by pivots at A, B, C, and D. HK is a beam rigidly attached to AC, and LM is a beam rigidly attached to BD; the angles AHK and BLM being right angles. A weight P is hung at K and a weight W is hung at M: it is required to find the ratio of P to W when there is equilibrium, neglecting the weights of the beams.

Since EF is vertical so are AC and BD; and KH and LM are horizontal.

The piece formed of KH and AC is acted on by the weight P, by a force at A arising from the beam AB, and by a force at C arising from the beam CD. Resolve the force at A into two components, one vertical and the other in the line AB. Resolve the force at C into two components, one vertical and the other in the line CD.

The components in the lines AB and CD must be equal and unlike; for if they were not the sum of the horizontal components of the forces on the piece would not be zero.

Let Y denote the vertical force on the piece at A, supposed upwards: then the vertical force on the piece at C must be P- Y upwards: for if it were not the sum of the vertical components of the forces on the piece would not be zero.

Thus AB is acted on at A by some force in the line AB, and by a vertical force Y downwards. And CD is acted on at C by some force in the line CD, and by a vertical force P- Y downwards.

Similarly AB is acted on at B by some force in the line AB, and by a vertical force downwards which we may denote by Z. And CD is acted on at D by some force in the line CD, and by a vertical force W-Z downwards.

Also AB is acted on by some force at E, and CD by some force at F.

Take moments round E for AB: thus

therefore

YX EA sin EAC-Z× EB sin EBD;

Yx EA=Z× EB.

Take moments round F for CD; thus

(P-Y) FC sin FCA=(W-Z) FD sin FDB;

therefore

(P—Y) EA=(W−Z) EB.

Hence, by addition,

PEA = W × EB.

Thus the ratio of P to W is independent of the lengths of HK and LM; and if EA= EB then P= W.

T. M.

13

Miscellaneous Examples in Statics.

1. The magnitudes of two bodies are as 3 is to 2, and their weights are as 2 is to 1: compare their densities.

2. Two forces act on a particle in directions at right angles to each other; they are in the ratio of 5 to 12, and their resultant is equal to 65 lbs.: find the forces.

3. Three forces represented by 24, 25, and 7 are in equilibrium when acting on a particle: shew that two of them are at right angles.

4. The resultant of two forces which act at right angles on a particle is 51 lbs.; one of the components is 24 lbs.: find the other.

5. Two forces acting in opposite directions to one another on a particle have a resultant of 28 lbs.; and if they acted at right angles they would have a resultant of 52 lbs.: find the forces.

6. ABC is a triangle and D the middle point of BC; three forces represented in magnitude and direction by AB, AC, DA act on a particle at A: find the magnitude and direction of the resultant.

7. Three forces 3, 4, 5 act on a particle in the centre of a square in directions towards three of the angles of the square: find the magnitude and direction of the force which will keep the particle at rest.

8. A and B are fixed points; PA and PB represent forces; if P moves along a fixed straight line, shew that the extremity of the straight line representing the resultant of the two forces will move along another fixed straight

9. Three forces, represented by those diagonals of three adjacent faces of a cube which meet, act at a point: shew that the resultant is equal to twice the diagonal of the cube.

10. A string passing round a smooth peg is pulled at each end by a force equal to 10 lbs., and the angle between the parts of the string on opposite sides of the peg is 120°: find the pressure on the peg, and the direction in which it acts.

11. Three smooth pegs are fastened in a vertical plane so as to form an isosceles triangle with the base horizontal and the vertex downwards, and vertical angle equal to 120°. A fine string with a weight W attached to each end is passed under the lower peg and over the other two. Find the pressure on each peg. Find also the vertical pressure on each peg.

12. Find a point within an equilateral triangle at which if a particle be placed it will be kept in equilibrium by three forces represented by the straight lines joining the point with the angular points of the triangle.

13. Forces represented in magnitude and direction by the diagonals of a parallelogram act at one of the angles: find the single force which will counteract them.

14. If R be the resultant of two forces P and Q acting on a particle, and S the resultant of P and R, shew that the resultant of S and Q will be 2R.

15. Three equal forces act at a point, and their directions are parallel to three consecutive sides of a regular hexagon: find the magnitude and the direction of the resultant.

16. Shew that if one of two forces acting on a particle be given in magnitude and position, and also the direction of their resultant, the locus of the extremity of the straight line representing the other will be a straight line.