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(19) Two equal weights P, P, attached to the ends of a string, which passes over two pegs A, B, fig. (32), in a horizontal line, sustain a weight W suspended at the end of a string the higher end of which is fixed to a point in the former string: find the magnitude of the angle AOB, when there is equilibrium, and ascertain whether it is possible for O to lie in the line AB.
If AOB=20, then cos 0 = a result which shews that, unless P be infinite, O must lie below the line AB.
(20) A weight W, fig. (33), is sustained by a string DW, the upper end of which is looped to a string ABDCA, which hangs over three tacks A, B, C, fixed at the angles of an equilateral triangle, B and C being in a horizontal line: find the pressure on the tacks, the portion BDC of the string being supposed to be equal to the portion BAC.
The pressure on A is equal to W, and that on B or C
(21) A string, of length c, passes over four tacks forming a square ABCD, fig. (34), of which two sides are horizontal, the length of each side being a: a weight W is suspended by a string hanging from the former one by a loop at E: find the tension of the former string.
(22) A string, the extremities of which are fastened to the ends of a uniform bar of known weight, passes over four tacks so as to form with the bar a regular hexagon, the bar being hori
zontal: find the vertical component of the pressure upon either of the two highest tacks.
The required vertical component is equal to half the weight of the bar.
SECT. 3. Components along any two lines not parallel.
(1) To prove that three equal forces, represented by OA, OB, OC, fig. (35), which act on a point 0, may be balanced by a single force LO, L being the point of intersection of the perpendiculars from the angles of the triangle ABC on the opposite sides.
The sum of the components of the three equal forces OA, OB, OC, parallel to BC, is represented by the projection of OA upon BC: but this projection is equal to that of OL, and therefore the sum of the components of OA, OB, OC, parallel to BC is equal and opposite to the component of LO parallel to this same line.
The analogous proposition is true in regard to either CA or AB.
Hence the four forces OA, OB, OC, LO, will be in equilibrium.
COR. The resultant of the three equal forces OA, OB, OC, is represented in magnitude and direction by OL.
SECT. 4. Friction.
(1) If the roughness of a plane, which is inclined to the horizon at a known angle, be such that a body will just rest on it, to find the least force along the plane requisite to drag the body up.
Let a be the inclination of the plane, W the weight of the body, R the reaction of the plane, P the required force, μ the
coefficient of friction. Then, resolving forces along and perpendicularly to the plane, we have
P=μR+W sin a,
R = W cosa,
P= (u cos a + sin a) W.
But, since the body just rests on the plane without support, μtan a: hence
P=2W sin ɑ.
(2) A body is supported on a rough inclined plane by a force acting along the plane: supposing the greatest magnitude of the force to be double the least magnitude, to determine the inclination of the plane to the horizon.
Let W be the weight of the body, R the reaction of the plane; P the least magnitude of the force, and accordingly 2P the greatest. Let μ represent the coefficient of friction and a the inclination of the plane. The forces producing equilibrium in the two cases are exhibited in the figures (36) and (37).
For equilibrium, in the first case, we have, resolving forces parallel and perpendicularly to the plane,
The corresponding equation in relation to the second figure, replacing P by 2P and μ by — μ, is
2P-μ W cos a =
W sin a....
Multiplying (1) by 2, and subtracting (2) from the resulting
equation, we have
3μ W cos a = W sin a,
tan a =
a = tan1 (3μ).
A heavy body is kept at rest on a given inclined plane by a force making a given angle with the plane; to prove that the
reaction of the plane, when it is smooth, is an harmonic mean between the greatest and least normal reactions, when it is rough.
Let W be the weight of the body, P the supporting force, R the normal reaction, AR the frictional reaction estimated up the plane; also let a be the inclination of the plane, and e the inclination of P to the plane.
Then, resolving along and at right angles to the plane, we have
Pcos e +λR= W sin a,
P sin e+ R W cos a.
Eliminating P between these equations, we have
R (cos e-λ sine) = W cos (a + e),
Hence, μ being the coefficient of friction, and R', R", the values of R when λ is equal +μ, μ, respectively, which are the extreme limits of the value of X,
COS € + μ sin Ε
W cos (a+e) ' R" ̄ W cos (a + €) '
R, being the reaction when the plane is smooth.
(4) P is the lowest point on the rough circumference of a circle, in a vertical plane, at which a particle P can rest, the coefficient of friction being unity: determine the inclination of the radius through P to the horizon.
(5) A given horizontal force Pjust supports a weight W on a rough plane inclined to the horizon at an angle : the same
weight will just rest without support on a plane of the same material, when the inclination is a determine 0.
The angle is given by the equation
(6) A weight is to be conveyed to the top of a rough plane, inclined to the horizon at an angle a: prove that, if the coefficient of friction be greater than
sec a- tan a,
it will be easier to lift the weight than to drag it up by means of a cord parallel to the plane.
(7) If the ratio of the greatest to the least force, acting parallel to a given rough inclined plane, which will support a certain weight on the plane, be equal to the ratio of the weight to the normal pressure on the plane, determine the coefficient of friction.
If a be the inclination of the plane to the horizon, the coefficient of friction is equal to
(8) A body is just supported on a rough inclined plane by a force acting along it; find the angle between this plane and a smooth plane, on which the body may be supported by the same force acting at the same angle to the horizon as before.
If μ denote the coefficient of friction, the required angle is equal to tan (u).
(9) A particle, the weight of which is W, rests upon a rough horizontal table, and is acted upon by two horizontal forces m W and n W, the directions of which are at right angles to each other; find the least value of the coefficient of friction consistent with the equilibrium of the particle.
The least value of the coefficient of friction is equal to
(m2 + n2)3.