By George E. Drabble
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We must now develop this technique to find the resultant of more than two forces. (5) The Polygon of Forces One fears proportionate complication when we increase the number of forces; fortunately the added complication is one of number, not of principle. Look at the five forces shown in Fig. 14. Let us assume that they are in equilibrium. Now we know that if we have only three forces, we can prove equilibrium by drawing the triangle of forces. But we now have a method of reducing two forces to one, by obtaining the resultant.
Since there is only the one force, the resultant acceleration a2 must be in the direction of this force, and is so shown. The new equation of motion is: W = ma2 and substituting as before: mg = ma2 from which a2 is clearly g. ) We can again use the equation: & = u1 + lax In this case, v is zero (as the shell comes to rest instantaneously at the top of its flight); u is the initial velocity, which we have just worked out (as v for the first phase of motion); a is a2 = g; and x is the height we wish to find.
P is given. W is obtained from the product of m and g. As usual, the 50 Applied Mechanics Made Simple value of g may be taken as 9-81 metres per second per second. However, substituting mg for W, and rearranging: mat mg Therefore P _ m g ai 3 x 106 -9-81 200 = 15 0 0 0 - 9 - 8 1 =^= 15 000 metres per second per second The value of 9-81 is so small that we can neglect it in this calculation. In physical terms, the weight of the shell is so small in comparison with the Λ Λ 02 |1 I w w (a) (b) Fig.
Applied Mechanics. Made Simple by George E. Drabble