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

A uniform rod having a weight of 35 N is resting horizontally on two supports 𝐴 and 𝐵 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 𝐴, determine the reactions of the two supports 𝑅_(𝐴) and 𝑅_(𝐵).

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### Video Transcript

A uniform rod having a weight of 35 newtons is resting horizontally on two supports 𝐴 and 𝐵 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 𝐴, determine the reactions of the two supports 𝑅 𝐴 and 𝑅 𝐵.

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 we’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, 𝑅 𝐴 and 𝑅 𝐵. These will act vertically upwards. We’re told that the distance between the two supports 𝐴 and 𝐵 is 48 centimetres. There is a weight of magnitude 24 newtons suspended at a point on the rods. We’re told that this point is 38 centimetres from 𝐴. The distance from 𝐴 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, 𝑅 𝐴 and 𝑅 𝐵. We also have two forces going downwards, 35 newtons and 24 newtons. 𝑅 𝐴 plus 𝑅 𝐵 is equal to 35 plus 24. 35 plus 24 is equal to 59. Therefore, 𝑅 𝐴 plus 𝑅 𝐵 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 𝐴 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 𝐴, whereas 𝑅 𝐵 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 𝑅 𝐵 is equal to 𝑅 𝐵 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 𝑅 𝐵. 840 plus 912 is equal to 1752. Dividing both sides of this equation by 48 gives us 𝑅 𝐵 is equal to 36.5. We can now substitute this value into equation one. 𝑅 𝐴 plus 36.5 is equal to 59. Subtracting 36.5 from both sides gives us 𝑅 𝐴 is equal to 22.5. The reactions of the two supports are 𝑅 𝐴 equals 22.5 newtons and 𝑅 𝐵 equals 36.5 newtons.