By Brian H. Chirgwin

**Read Online or Download A Course of Mathematics for Engineers and Scientists. Volume 3: Theoretical Mechanics PDF**

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**Extra resources for A Course of Mathematics for Engineers and Scientists. Volume 3: Theoretical Mechanics **

**Example text**

3 (ii)] and an equal parallel force ' at A is equivalent to ]ACA + 12AB, in position as well -1-i1CB at A. But 12CB as magnitude, since all three forces pass through A. •. AEF = 1A0B — BC CA). Similarly yFE = ly(BC — CA + AB), vDE_ v (C A — AB+ B Hence the equivalent set of forces is v — A)BC i(v + A — y)CA y — v)AB. Exercises 2:2 1. Forces of equal magnitude F act at opposite ends of the diameter x = 0 of the light circular disc of boundary x2 + y2 = a2 and in its plane. The force at (0, — a) is parallel to the negative x-axis, and the force at (0, a) makes an angle 0(0 < 0 < n/2) with the x-axis.

Find the tension in the bar CD. The tension T, in CD is an internal force so we must violate the condition that CD is fixed in length in order to find T. , AB, BC, CA etc. are all fixed in length and B and E remain on the same level (Fig. 7). We therefore consider the framework in which < DAC = 0,

The frame is kept in shape by two light struts AC, AD. Find the tension in the bar CD. The tension T, in CD is an internal force so we must violate the condition that CD is fixed in length in order to find T. , AB, BC, CA etc. are all fixed in length and B and E remain on the same level (Fig. 7). We therefore consider the framework in which < DAC = 0,