WEBVTT
00:00:00.540 --> 00:00:14.720
A uniform rod having a weight of 35 newtons is resting horizontally on two supports π΄ and π΅ at its ends, where the distance between the supports is 48 centimetres.
00:00:15.450 --> 00:00:28.800
If a weight of magnitude 24 newtons is suspended at a point that is 38 centimetres away from π΄, determine the reactions of the two supports π
π΄ and π
π΅.
00:00:30.690 --> 00:00:37.660
In order to solve this problem, we will resolve vertically and also take moments about a point.
00:00:38.860 --> 00:00:43.890
Before doing this, it is worth drawing a diagram with all the forces shown.
00:00:44.880 --> 00:00:51.670
As weβre told the rod is uniform, the weight of 35 newtons will act at the centre of the rods.
00:00:52.550 --> 00:00:58.330
At either end of the rod, there will be two reaction forces, π
π΄ and π
π΅.
00:00:59.200 --> 00:01:01.880
These will act vertically upwards.
00:01:03.050 --> 00:01:09.290
Weβre told that the distance between the two supports π΄ and π΅ is 48 centimetres.
00:01:10.260 --> 00:01:16.050
There is a weight of magnitude 24 newtons suspended at a point on the rods.
00:01:17.170 --> 00:01:21.730
Weβre told that this point is 38 centimetres from π΄.
00:01:23.040 --> 00:01:31.900
The distance from π΄ of the 35-newton force will be 24 centimetres, as a half of 48 is 24.
00:01:33.050 --> 00:01:40.630
We can now set up our two equations, firstly, by resolving vertically and then taking moments about a point.
00:01:41.760 --> 00:01:49.940
As the system is in equilibrium, the forces going vertically upwards will be equal to the forces going vertically downwards.
00:01:50.530 --> 00:01:55.180
We have two forces going upwards, π
π΄ and π
π΅.
00:01:56.150 --> 00:02:02.240
We also have two forces going downwards, 35 newtons and 24 newtons.
00:02:02.960 --> 00:02:08.420
π
π΄ plus π
π΅ is equal to 35 plus 24.
00:02:09.620 --> 00:02:12.980
35 plus 24 is equal to 59.
00:02:13.270 --> 00:02:17.190
Therefore, π
π΄ plus π
π΅ is equal to 59.
00:02:17.950 --> 00:02:20.350
We will call this equation one.
00:02:21.710 --> 00:02:24.410
We will now take moments about a point.
00:02:24.720 --> 00:02:31.310
And in this case, we will choose point π΄ as we know all the distances from this point.
00:02:32.530 --> 00:02:40.120
The moment of any force about a point is equal to the force multiplied by its distance from the point.
00:02:41.470 --> 00:02:51.400
We also know that as the rod is in equilibrium, the sum of the moments in a clockwise direction are equal to the sum of the moments in the anticlockwise direction.
00:02:53.010 --> 00:03:03.340
The 35-newton force and 24-newton force act clockwise around π΄, whereas π
π΅ acts anticlockwise.
00:03:04.710 --> 00:03:11.140
The moment of the 35-newton force is equal to 35 multiplied by 24.
00:03:12.300 --> 00:03:18.450
The moment of the 24-newton force is equal to 24 multiplied by 38.
00:03:19.300 --> 00:03:25.420
Finally, the moment of π
π΅ is equal to π
π΅ multiplied by 48.
00:03:26.210 --> 00:03:30.980
35 multiplied by 24 is equal to 840.
00:03:31.940 --> 00:03:36.580
24 multiplied by 38 is equal to 912.
00:03:37.610 --> 00:03:43.490
840 plus 912 is equal to 48 π
π΅.
00:03:44.620 --> 00:03:50.730
840 plus 912 is equal to 1752.
00:03:51.580 --> 00:03:59.000
Dividing both sides of this equation by 48 gives us π
π΅ is equal to 36.5.
00:04:00.110 --> 00:04:04.120
We can now substitute this value into equation one.
00:04:05.050 --> 00:04:09.290
π
π΄ plus 36.5 is equal to 59.
00:04:10.020 --> 00:04:17.670
Subtracting 36.5 from both sides gives us π
π΄ is equal to 22.5.
00:04:18.790 --> 00:04:29.590
The reactions of the two supports are π
π΄ equals 22.5 newtons and π
π΅ equals 36.5 newtons.