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EXAMPLES.

61. (i) Let AB be an inclined plane hinged at A, so that B can move in a circular arc and suppose that B can be fixed at any desired point.

Suppose AB is originally horizontal and that B is gradually raised until the slider is just on the point of slipping.

Up to the slipping point, the weight, the friction, and the reaction form three forces in equilibrium.

F

E

W

Fig. 39.

Let OD represent the weight W, then drawing through the ends of OD, OE parallel to AB and DE perpendicular to it, we find

EO = F the friction,

DER the reaction.

This construction will give the friction and reaction for any position of AB before slipping takes place.

For the greatest possible inclination of AB, the value of F will be greatest and will then be the limiting friction.

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(ii) A body of weight W on a given inclined plane is acted on by a force parallel to the plane, required to find the force so that the body will just be on the point of moving down the plane, the coefficient of friction being supposed known.

Draw the vector polygon ODE as before, and along EO set off EF=μ.DE; then FO is the force required.

If EF >EO, then the force P=F0 must act down the plane.

Suppose μ; then ED must be divided into 8 equal parts and EF equals one of these.

(iii) As a further example consider the case of a body just kept from sliding down a plane by a horizontal force, to determine this force and the reaction of the plane.

At first sight this appears difficult since neither P, R nor F is known, but we are supposed to know the coefficient of friction μ

W

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F=μR.

Hence draw two lines in proper directions to represent R and μR to some unknown scale; through the points A and B thus determined draw AC vertical and BC horizontal. Then AC represents W and since W is known the scale to which the drawing is made can be determined.

R

If the force P be applied horizontally then we can see from the figure at what Bangle to apply it so that P may have the least possible value. For the ratio of P to W is given by the ratio of BC to CA and this will be least when BC is perpendicular to AB.

P
Fig. 40.

62. Theorem: If three non-parallel forces be in equilibrium they must all pass through a common point. (The case of parallel forces will be considered in the chapter on Rotors.)

Consider any two of them P and Q intersecting at O, then their resultant R passes through O and hence the equilibrant or third force must also pass through O. If not, then two forces not in the same straight line may be in equilibrium, which is contrary to experience.

C

63. This Theorem is of great use in finding the direction and

A

Fig. 41.

M

R

magnitude of the unknown forces with which two bodies press on one another when in contact.

For instance, suppose AB is a uniform plank hinged at A and that the other end B is fastened by means of a cord or light chain to some fixed hook at C, and we require to find the tension in the chain and the force (reaction) with which the bolt at A acts on the plank.

The string can only exert a force in the direction of its length, and assuming that the weight of the body

acts at the mid-point M of the plank we get by the intersection of the two lines CB and MD the point D, through which the reaction R of the bolt must pass.

E

R

W

F

Knowing W the weight of the plank, set off a length vertically downwards representing it to scale, through the ends of this vector draw lines parallel to AD and CD giving the closed vector polygon EFGE. Hence FG gives the magnitude and sense of the force at B, GE the magnitude and sense of the force at A. As a second example take the case of a ladder resting on a rough ground and against a smooth vertical wall.

Fig. 42.

Since the wall is smooth the reaction there must be perpendicular to the wall. Hence again we can find the point of intersection of two of the forces (viz. the weight of the ladder acting through the mid-point, and the reaction of the smooth wall) and can therefore find by the Vector Polygon the magnitudes and senses of the reactions where the ladder touches the wall and the ground.

EXERCISES IV.

(1) Forces of 10 lbs. and 15 lbs. respectively act at a point; find the resultant when the angle between them is

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(2) The forces acting on a body at a point are

10 lbs. N., 2lbs. N.E., 3 lbs. 30° N. of E., 7 lbs. W., 15 lbs. S.E. Find the resultant and the force which with the others will maintain equilibrium.

(3) The length of an inclined plane is 12 ft., its height is 3 ft.; find the horizontal force required to keep a weight of 15 lbs. in equilibrium on the plane (supposing no friction).

(4) In (3) find the force if it is inclined at an angle of 45° with the horizontal.

(5) A string fixed to a point A, passes over a smooth pulley at

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B and supports a weight of 7 lbs. at its end. A smooth ring carrying 12 lbs. weight can slide on the string; find the tension in the string and the vertical distance of O below AB, if AB is horizontal and 2 ft. in length.

(6) Four forces acting at a point are in equilibrium, one of 3 lbs. acts vertically downwards, one of 1.2 lbs. acts horizontally (say from left to right), the third makes an angle of 30° with the horizontal and the fourth is perpendicular to the third. Find the third and fourth forces.

(7) Two smooth inclined planes have a common vertex, a string passing over a smooth pulley at the vertex is attached to two weights 10 and 15 lbs. on the planes. The inclination of the plane with the 10 lb. load is 30°; find the reactions on the planes, the tension in the string, and the inclination of the other plane.

(8) Two weights of 3 lbs. are connected by a string and suspended over two smooth pulleys on the same level 2 ft. apart. A weight of 4 lbs. is then suspended from the string; find the tension of the string and the position of equilibrium.

(9) A body is just on the point of sliding up an inclined plane by the action of a horizontal force; find this force and the reaction of the plane, given that the inclination of the plane is 23°, the weight of the body being 10 lbs. and the coefficient of friction 3.

(10) A body of weight 10 lbs. is kept at rest on an inclined plane by a force parallel to the plane; if coefficient of friction=2 and the plane rises 1 in 10, find the greatest force that can be applied to the body without moving it (1) up the plane, (2) down the plane.

(11) A uniform beam of weight 500 lbs. is hinged at one end, the other rests on a smooth inclined plane of angle 20°; find the reactions at the hinge and the plane, the length of the beam being 20 ft. and the end in contact with the plane being 3 ft. higher than the hinge.

A

(12) A rectangular

B

trap-door is kept open by a cord applied at the centre of the edge opposite the hinge. The weight of the door is 50 lbs. and it is inclined to the horizon at an angle of 35°; find the tension in the cord

(1) when horizontal,

(2) when inclined at an angle of 40°, and determine the reaction of the hinge.

CHAPTER II.

MASS-CENTRES.

64. Mean Vector. By the mean of a set of arithmetical or algebraical quantities is meant the sum of the quantities divided by their number. Thus, the mean of a, b, c and d, is

a+b+c+d
4

The mean of a number of vectors is defined in a similar

manner.

Definition. A number of vectors being given, their sum divided by the number of vectors is called the mean of the given vectors. It is of course itself a vector.

If a, B, y, d,... be the given vectors, n their number and μ their mean, then is

or

μπ

a + B + y + 8+...

n

ημ= α + β +y+ 8+ ...

The sum of a number of similar quantities is often denoted by the Greek letter Σ with the first of the given quantities written immediately after it. So that the above equations are written in the abbreviated forms

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