49. [Fig. 19] represents a wooden rod 4' long, sustained by resting on two supports a and b, and having the length a b divided into 14 equal parts. Let a weight of 14 lbs. be hung on the rod at its middle point c; this weight must be borne by the supports, and it is evident that they will bear the load in equal shares, for since the weight is at the middle of the rod there is no reason why one end should be differently circumstanced from the other. Hence the total pressure on each of the supports will be 7 lbs., together with half the weight of the wooden bar.
50. If the weight of 14 lbs. be placed, not at the centre of the bar, but at some other point such as d, it is not then so easy to see in what proportion the weight is distributed between the supports. We can easily understand that the support near the weight must bear more than the remote one, but how much more? When we are able to answer this question, we shall see that it will lead us to a knowledge of the composition of parallel forces.
PRESSURE OF A LOADED BEAM
ON ITS SUPPORTS.
51. To study this question we shall employ the apparatus shown in [Fig. 20]. An iron bar 5' 6" long, weighing 10 lbs., rests in the hooks of the spring balances a, c, in the manner shown in the figure. These hooks are exactly five feet apart, so that the bar projects 3" beyond each end. The space between the hooks is divided into twenty equal portions, each of course 3" long. The bar is sufficiently strong to bear the weight b of 20 lbs. suspended from it by an S hook, without appreciable deflection. Before the weight of 20 lbs. is suspended, the spring balances each show a strain of 5 lbs. We would expect this, for it is evident that the whole weight of the bar amounting to 10 lbs. should be borne equally by the two supports.
52. When I place the weight in the middle, 10 divisions from each end, I find the balances each indicate 15 lbs. But 5 lbs. is due to the weight of the bar. Hence the 20 lbs. is divided equally, as we have already stated that it should be. But let the 20 lbs. be moved to any other position, suppose 4 divisions from the right, and 16 from the left; then the right-hand scale reads 21 lbs., and the left-hand reads 9 lbs. To get rid of the weight of the bar itself, we must subtract 5 lbs. from each. We learn therefore that the 20 lb. weight pulls the right-hand spring balance with a strain of 16 lbs., and the left with a strain of 4 lbs. Observe this closely; you see I have made the number of divisions in the bar equal to the number of pounds weight suspended from it, and here we find that when the weight is 16 divisions from the left, the strain of 16 lbs. is shown on the right. At the same time the weight is 4 divisions from the right, and 4 lbs. is the strain shown to the left.
Fig. 20.
53. I will state the law of the distribution of the load a little more generally, and we shall find that the bar will prove the law to be true in all cases. Divide the bar into as many equal parts as there are pounds in the load, then the pressure in pounds on one end is the number of divisions that the load is distant from the other.
54. For example, suppose I place the load 2 divisions from one end: I read by the scale at that end 23 lbs.; subtracting 5 lbs. for the weight of the bar, the pressure due to the load is shown to be 18 lbs., but the weight is then exactly 18 divisions distant from the other end. We can easily verify this rule whatever be the position which the load occupies.
55. If the load be placed between two marks, instead of being, as we have hitherto supposed, exactly at one, the partition of the load is also determined by the law. Were it, for example, 3·5 divisions from one end, the strain on the other would be 3·5 lbs.; and in like manner for other cases.