# Question Video: Finding the Reactions of the Supports of a Rod in Equilibrium Mathematics

A uniform rod having a weight of 35 N is resting horizontally on two supports π΄ and π΅ at its ends, where the distance between the supports is 48 cm. If a weight of magnitude 24 N is suspended at a point that is 38 cm away from π΄, determine the reactions of the two supports π_(π΄) and π_(π΅).

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### Video Transcript

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. 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΅.

In order to solve this problem, we will resolve vertically and also take moments about a point. Before doing this, it is worth drawing a diagram with all the forces shown. As weβre told the rod is uniform, the weight of 35 newtons will act at the centre of the rods. At either end of the rod, there will be two reaction forces, π π΄ and π π΅. These will act vertically upwards. Weβre told that the distance between the two supports π΄ and π΅ is 48 centimetres. There is a weight of magnitude 24 newtons suspended at a point on the rods. Weβre told that this point is 38 centimetres from π΄. The distance from π΄ of the 35-newton force will be 24 centimetres, as a half of 48 is 24.

We can now set up our two equations, firstly, by resolving vertically and then taking moments about a point. As the system is in equilibrium, the forces going vertically upwards will be equal to the forces going vertically downwards. We have two forces going upwards, π π΄ and π π΅. We also have two forces going downwards, 35 newtons and 24 newtons. π π΄ plus π π΅ is equal to 35 plus 24. 35 plus 24 is equal to 59. Therefore, π π΄ plus π π΅ is equal to 59. We will call this equation one.

We will now take moments about a point. And in this case, we will choose point π΄ as we know all the distances from this point. The moment of any force about a point is equal to the force multiplied by its distance from the point. 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. The 35-newton force and 24-newton force act clockwise around π΄, whereas π π΅ acts anticlockwise.

The moment of the 35-newton force is equal to 35 multiplied by 24. The moment of the 24-newton force is equal to 24 multiplied by 38. Finally, the moment of π π΅ is equal to π π΅ multiplied by 48. 35 multiplied by 24 is equal to 840. 24 multiplied by 38 is equal to 912. 840 plus 912 is equal to 48 π π΅. 840 plus 912 is equal to 1752. Dividing both sides of this equation by 48 gives us π π΅ is equal to 36.5. We can now substitute this value into equation one. π π΄ plus 36.5 is equal to 59. Subtracting 36.5 from both sides gives us π π΄ is equal to 22.5. The reactions of the two supports are π π΄ equals 22.5 newtons and π π΅ equals 36.5 newtons.