Example i. A mass m pounds is placed under the action of gravity on a smooth inclined plane; find its acceleration. Let AB be the inclined plane; draw AC horizontal and BC vertical. Let the angle BAC be a. The weight of the mass P is a force mg poundals vertically downwards; let PN represent this force. Resolve this force into its two components PQ and PG, in the directions along and perpendicular to PA respectively. Then instead of mg poundais vertically downwards we have PQ=mg sin a poundals along PA and PG=mg cos a poundals along PG. Now the presence of the plane prevents any motion in the mass in the direction PG. Also, the plane is 'smooth’; by which is meant, that the stress between the mass and the plane must be perpendicular to the plane. These conditions will both be satisfied if we suppose that the plane exerts a pressure on the mass, in the direction GP, just sufficient to prevent the mass from having any motion in that direction. Hence we say, that the pressure of the plane on the mass in the direction GP is mg cos a poundals. Thus, the forces acting on the mass are 'mg sin a poundals along PA, mg cos a poundals along PG and – mg cos a poundals in the same direction. Therefore, the resultant force is mg sin a poundals along PA. sin a celos. Example ii. A mass m lbs. is placed on a smooth inclined plane and is acted on by its own weight and by a horizontal force p poundals ; find its acceleration. With the construction of Example i. resolve the weight and the force p poundals along and perpendicular to the plane. Case (i) We have if p poundals act as the forces in figure I. (mg sin a -p cos a) poundals, along PA, (mg cos a +p sin a) poundals - the pressure of the plane, along PG. с Assuming that the resultant acceleration is along AB, it follows that the resultant force must be along AB; so that the pressure of the plane on the mass=(mg cos a +p sin a) poundals; therefore the resultant force is (mg sin a -p cos a) poundals along PA, which produces in mass m lbs. the acceleration ( g sin a - celos along PA. If this acceleration happens to be zero the forces are in equilibrium; that is, if mg sin a=p cos a. Case (ii) if p act as in figure II., then as before, $a) ni B p+ u mg a с the pressure of the plane on the mass=(mg cos a - p sin a) poundals. And the acceleration along PA p a celos, (8$ sa m a EXAMPLES. XXIX. 1. Forces of 3 poundals and 4 poundals at right angles to each other act upon a mass of 5 lbs.; find the acceleration produced. How long will the mass take to move 18 ft. from rest? 2. Equal forces of 10 poundals at an angle of 120° to each other act on a mass of 1 lb.; how long will the mass take to acquire a velocity of 30 miles an hour? 3. A mass of m lbs. is let fall from rest under the action of gravity and is acted on by a horizontal force of 24 poundals; find how long it will take to fall a vertical distance of 144 ft.; and how far it will move horizontally in the same time. 4. Two horses, attached by ropes to a mass placed on a smooth horizontal plane and weighing i ton, pull at right angles to each other, one with a force equal to the weight of 3 cwt. and the other with a force equal to the weight of 4 cwt.; find the direction of motion and the acceleration of the mass. 5. If in Question 4 the plane is rough so that there is a force of friction which retards the motion equal to 15th of the pressure of the mass on the plane; find the acceleration. 6. A particle slides down a smooth inclined plane whose inclination to the horizon is 30°; shew that it takes twice as long to descend a given vertical distance as a particle falling freely under the action of gravity. 7. Forces of 2 poundals, 3 poundals and 1 poundal act along lines OA, OB, and OC so that the angle AOB=45o=BOC upon a mass of 4 lbs. ; find the acceleration produced. 8. Forces of 2 poundals, 3 poundals and 1 poundal act, along lines 3 0.4, OB and OC, so that the angle AOB=30°, BOC=15°, upon a mass of i lb.; find the acceleration produced. 9. Prove that the mass in Example 3 moves in a straight line. 10. A mass P is drawn up a smooth plane inclined at an angle of 30° to the horizon by means of an equal mass which descends vertically, the masses being connected by a string which passes over a small pulley at the top of the plane; prove that the acceleration is Àg celos. 11. Find the ratio of the masses of P and Q in Question 10 that they may have no acceleration. & celos. 12. Suppose that P in Question 10 starts from rest at the bottom of the plane and the interval t seconds occupied by Q in drawing P to the top of the plane is noticed ; then P and Q being interchanged the interval ť seconds occupied by P in drawing © up the plane is noticed ; shew that if t= 2t' then Q= P. 13. A mass m lbs. is drawn up a smooth plane inclined at an angle a to the horizon, by means of a mass m' lbs. which descends vertically, the masses being connected by a string which passes over a small pulley at the top of the plane ; prove that the acceleration is m sin a - ni' m + m 14. Two masses ni lbs. and m' lbs. are connected by a light string which passes over a small smooth pulley fixed at the top of two smooth inclined planes of equal heights placed back to back, the inclination of the planes being each 30°; shew that when the masses are equal they have no acceleration. 15. In Question 14 shew that when mi = 1 m the acceleration is tg celos. 16. If in Question 14 the inclinations of the planes to the horizon are a and a' respectively, prove that the acceleration is m sin a – m' sin a' -g celos, m + m' and that the tension of the string is mm' (sin a + sin a') 8 poundals. m + m' 17. Determine the acceleration due to gravity from the following experiment. A mass descending vertically under the action of gravity draws an equal mass a distance of 25 ft. from rest in 23 seconds up a smooth plane inclined at an angle 30° to the horizon, by means of a light string passing over a smooth small pulley at the top of the plane. Example iïi. To find the interval occupied by a particle in sliding durun a smooth inclined plane of length s ft. and inclination a. By Example i. the particle has an acceleration g sin a celos down the plane. The interval t seconds, occupied by a point in moving from rest over a distance of s ft. with acceleration g sin a celos, is given by s= (g sin a)ta, whence, 25_1 t= Example iv. A series of chords are drawn from the highest point of a fixed vertical circle; prove that the intervals occupied by a particle in sliding from rest down each of these chords (supposed smooth) are all equal. B Using the notation of Example iii, let ABC be the circle; AC one of the chords; CN horizontal; then AC=s ft. and ACN=a. The interval occupied in sliding down AC, by Example iii, is s 25 seconds. Is sin But AC=AB sin a, sin a That is, =AB=2r; the length of the radius being r feet. Hence, the interval in question is |