Elements of Mechanics: For the Use of Colleges, Academies, and High SchoolsA.S. Barnes & Company, 1866 - 331 pages |
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Page 108
... inches , that of the power , 33 , and the resistance 100 lbs . What is the power necessary to hold the resistance in equilibrium ? Ans . 8 lbs . 2. Four weights of 1 , 3 , 5 , and 7 lbs . respectively , are suspended from points of a ...
... inches , that of the power , 33 , and the resistance 100 lbs . What is the power necessary to hold the resistance in equilibrium ? Ans . 8 lbs . 2. Four weights of 1 , 3 , 5 , and 7 lbs . respectively , are suspended from points of a ...
Page 109
... inches . What are the weights , and their lever arms ? Ans . The weights are 7 lbs . and 3 lbs . ; their lever arms are 153 in . , and 63 in . 6. The apparent weight of a body weighed in one pan . of a false balance is 5 lbs . , and in ...
... inches . What are the weights , and their lever arms ? Ans . The weights are 7 lbs . and 3 lbs . ; their lever arms are 153 in . , and 63 in . 6. The apparent weight of a body weighed in one pan . of a false balance is 5 lbs . , and in ...
Page 123
... inches . 2. The radius of the axle of a windlass is 3 inches , and the crank - arm 15 inches . What power must be applied to the crank - handle , to support a resistance of 180 lbs . , applied to the circumference of the axle ? Ans . 36 ...
... inches . 2. The radius of the axle of a windlass is 3 inches , and the crank - arm 15 inches . What power must be applied to the crank - handle , to support a resistance of 180 lbs . , applied to the circumference of the axle ? Ans . 36 ...
Page 176
... inches long , branched towards its lower end , so as to embrace a cylindrical glass vessel 7 or 8 inches deep , and having 6.8 in . of this depth filled with mercury . The exact quantity of mercury being dependent on the weight and ...
... inches long , branched towards its lower end , so as to embrace a cylindrical glass vessel 7 or 8 inches deep , and having 6.8 in . of this depth filled with mercury . The exact quantity of mercury being dependent on the weight and ...
Page 178
... inches , when the force of gravity is reckoned at 321 ft ? Ans . 1.2387 sec . 4. How many vibrations will a pendulum 36 inches in 178 MECHANICS .
... inches , when the force of gravity is reckoned at 321 ft ? Ans . 1.2387 sec . 4. How many vibrations will a pendulum 36 inches in 178 MECHANICS .
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Common terms and phrases
acceleration due algebraic sum angular velocity atmosphere axes body called centre of gravity centrifugal force cistern coefficient coefficient of friction components cord cubic cubic foot curve cylinder denote the angle diameter diminished direction distance entire equal Equation feet fluid force acting force of gravity friction Hence horizontal hydrometer inches inclined plane inertia infinitely small initial velocity instrument length lever arm liquid machine mass mechanical advantage mercury moment of inertia moving force multiplied orifice parallelogram of forces particles pendulum perpendicular pipe piston point of application position power and resistance principle of moments projection pulley pump quantity radius represent reservoir respect rest resultant right angles rolling friction rope rotation scale-pan SOLUTION space described specific gravity square steam Substituting suppose temperature tension tion triangle tube upper surface upwards vertical vessel vibration volume weight whence
Popular passages
Page 271 - ... and altitude equal to the depth of the centre of gravity of the surface below the surface of the fluid.
Page 28 - Universal Gravitation. It is, therefore, subject to the same law, that- is, it varies directly as the mass of the body acted •upon, and inversely as the square of its distance from the centre of the earth.
Page 211 - The moment of inertia of an area with respect to any axis is equal to the moment of inertia with respect to a parallel axis through the...
Page 82 - ... y" z", being those of its extremities; whence we conclude that the centre of gravity of a straight line is at its middle point. Example 2. — Find the centre of gravity of the perimeter of a polygon. This may be done, according to Equations (90), by taking the sum of the products which result from multiplying the length of each side by the co-ordinate of its middle point, and dividing this sum by the length of the perimeter...
Page 56 - Hence, the moment of the resultant of two forces is equal to the algebraic sum of the moments of the forces taken separately. 53. Forces Acting at Different Points. Parallel Forces.— We have thus far considered forces acting upon a single particle, or upon one point of a body. If, how- Fia 33...
Page 212 - J., in reference to an axis through its centre of gravity and perpendicular to the plane containing this point and the normal. In what precedes, no reference is made to friction, but thus far no it is obvious that this principle cannot be wholly dis...
Page 258 - ... new ways to kill more and more people." Nations, not science, build fiendish machines to slaughter the citizens of other nations. The physical knowledge of matter and energy, even though essential to the invention of the atomic bomb, will not explode. The now famous equation E = me2 (the energy in a body is equal to its mass multiplied by the square of the speed of light) is in itself quite harmless. If the world's societies were dedicated to peace, the powerful forces predicted by that equation...
Page 318 - Archimedes stated that a body immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid.