By George E. Drabble

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Now, the upward-acting force Fx and the force of 2 units both exert a clockwise moment. All remaining moments are anti-clockwise. This gives our second equation thus: (Ft x 4) + (2 x 1) - (8 x 1) - (6 x 2) - (4 x 3) = 0 showing Fi to be 1\ kilonewtons. But we have done almost twice as much work as necessary. Since the whole beam is in equilibrium, we know that there can be no resultant force upwards or downwards. This by itself is not sufficient information to find Fi and F2, but having found one of these, we can use this fact to find the other.

Both cage and counterweight are each subjected to weight (downwards) and rope tension (upwards) only, if the frictional resistance of the air is ignored. The air resistance, which was one of the major forces in the first example, is here so small as not to seriously affect the problem. The two rope tensions have been f~\ -*-T w w (a) (b) (0 Fig. 9. Three examples of bodies subjected to forces, and the free-body diagrams: (a) an aircraft in flight; (b) a lift cage with counterweight; (c) a bullet in vertical flight.

In Chapter Three we have briefly studied force in its most general aspect, but have not considered the effects of forces. A considerable part of the remainder of this book will deal with the effects of forces in various situations. Chapters Eight to Ten will deal with the behaviour of some engineering materials when subjected to forces. In Chapters Eleven and Twelve we shall examine the forces encountered in static and moving fluids. But this chapter is concerned with one of the most fundamental aspects of dynamics—the effect of forces on movable bodies.

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