[Let P be the force acting parallel to the plane which is able to support the weight. Then 20x7660-20 × 6 × 6428=P, P-7.6064 lbs. If P' force parallel to base of plane, then = 7·6064=P'(·6428+6×·7660), .. P' 6.9 lbs. Pressure on plane=20×6428+P′ × ·7660=184 lbs. 40. A flat body is projected with a velocity of 20 feet per second in the direction of greatest slope upwards on an inclined plane of 1 in 3. The coefficient of friction being, find (1) how long it will continue rising, (2) how far up the plane it will reach, (3) how long till it passes again through the point of projection, and (4) with what velocity. Thus time till it passes again through point of projec tion=3.75 sec. (4) V-at= 32 x 2.65/2-14142) 2×2 (-1*1)= 8.28 feet per second.] 16 x 2.65 × 5858 3 41. A ladder, the centre of inertia of which may be supposed to be at the middle of its length, rests on a rough horizontal plane and against a rough vertical wall. Calling the coefficient of friction between the ground and the ladder μ, and that of the friction. between the ladder and wall u', find the greatest inclination of the ladder to the wall at which the ladder can rest. [Let be angle it makes with horizontal plane, R the reaction of the plane, and s the reaction of the wall. Then Taking moments about lower point of ladder, W cos 0=2R cos 0-2μR sin 0, 42. A piece of oak rests on the horizontal surface of an oak table. The resultant of gravity and other forces act upon it in a line at an angle of 27° to the vertical. Taking the coefficient of friction of oak on oak as 0.418; find whether the piece of oak will begin to slide or not. or or [We are to find whether g sin i> or < μg cos i, tan i> or < μ, •5095> or < 418; and since it is greater the oak will begin to slide.] CENTRES OF INERTIA. 43. Define Centre of Inertia. Prove that if x, y are the co-ordinates of the centre of inertia of a set of particles whose masses are m1, m„ m ̧, etc., and coordinates xy, x12; xy, etc. 2 m1x2+m2 x2+etc. m1y2+my+etc. ; y = m, +m2+ etc. 44. A hole 10 inches in diameter is cut in a uniform circular plate of 30 inches diameter. The centres of the circles are 5 inches apart. Find how far from the centre of the larger circle is the centre of inertia of the remaining part. [Take the centre of the larger circle as the origin of co-ordinates, and the line passing through the two centres as the negative axis of x, then. 45. From a circular disc a square has been cut. The middle point of one side of the square is at the centre of the disc, and the side of the square is equal to half the radius. Find the centre of inertia of the area that is left. [Take the centre of the circle as the origin of coordinates and the axis of x perpendicular to the side of the square which passes through the origin. Then if x denote the coordinate of the centre of inertia, 46. From the corner of an equilateral triangle another equilateral triangle is cut off the length of whose side is one third of the original triangle. Find the centre of inertia of the remainder. [Take the middle point of the base as the origin of co-ordinates and the line passing through this point and the vertex as the axis of y. Then if y denote the co-ordinate of the centre of gravity and a the perpendicular distance from vertex to base, we have 47. Find a formula for the velocity v of a rifle bullet fired into a ballistic pendulum from the following data:-Mass of pendulum 2500 grammes. The centre of inertia, the line of motion of the bullet, and the point of attachment of the tape are respectively 94, 100, 110 centimetres below the axis of suspension. 1.2 seconds is the time of oscillation of the equivalent simple pendulum, s the length of tape drawn out, and m the weight of the bullet in grammes. [Let M=mass of pendulum when bullet is lodged in it. Then taking moments of momentum round axis of suspension mvp-Mk2w, where p=100 cms. Again the potential energy of the pendulum in its position of greatest deflection, if is corresponding |