Question Video: Finding the Magnitude of the Weight of a Rod and the Distance between Its Line of Action and the End of the Rod Mathematics

𝐴𝐡 is a nonuniform rod having a length of 77 cm resting in a horizontal position on one support, which is 26 cm away from the end 𝐴. It is kept in equilibrium by suspending a weight of 16 N at its end 𝐴 and a weight of 2 N at the end 𝐡. If the distance between the support and the end 𝐴 is changed to be 23 cm, the rod will be kept in equilibrium by suspending a weight of 18 N at the end 𝐴 only. Find the magnitude of the weight of the rod π‘Š and the distance π‘₯ between its line of action and point 𝐴. Round your answers to the nearest two decimal places.

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Video Transcript

𝐴𝐡 is a nonuniform rod having a length of 77 centimeters resting in a horizontal position on one support, which is 26 centimeters away from the end 𝐴. It is kept in equilibrium by suspending a weight of 16 newtons at its end 𝐴 and a weight of two newtons at the end 𝐡. If the distance between the support and the end 𝐴 is changed to be 23 centimeters, the rod will be kept in equilibrium by suspending a weight of 18 newtons at the end 𝐴 only. Find the magnitude of the weight of the rod π‘Š and the distance π‘₯ between its line of action and point 𝐴. Round your answers to the nearest two decimal places.

In this question, we’re given two different scenarios about the rod 𝐴𝐡. We will begin by sketching both of these. We are told that rod 𝐴𝐡 is 77 centimeters long. In the first scenario, it is kept in a horizontal position by a single support which is 26 centimeters from the end 𝐴. This support will result in a normal reaction force 𝑅 acting vertically upwards. We are told that the rod is kept in equilibrium by suspending weights of 16 newtons and two newtons at ends 𝐴 and 𝐡, respectively.

In the second scenario, the support is placed 23 centimeters from end 𝐴. This time, it is kept in equilibrium by suspending a weight of 18 newtons at end 𝐴 only. We have been asked to calculate the magnitude of the weight of the rod π‘Š and the distance π‘₯ between its line of action and point 𝐴. It is important to note at this point that as 𝐴𝐡 is a nonuniform rod, the weight will not act at the midpoint, whereas if it was uniform, the weight would act at the midpoint of 𝐴𝐡.

We will answer this question using our knowledge of moments. When a body is in equilibrium, as in this case, we know that the sum of the moments equals zero, where the moment of a force is equal to 𝐹 multiplied by 𝑑. 𝐹 is a force acting at a point and 𝑑 is the perpendicular distance from this point to the point at which we’re taking moments. We will now clear some space and then take moments about a point in both scenarios.

We recall we are trying to find the magnitude of the weight of the rod π‘Š and the distance π‘₯ between its line of action and point 𝐴, giving our answers to two decimal places. We can take moments about any point on the rod, and this is often done at one of the ends. However, in this case, as we don’t know the value of the reaction force 𝑅, we will take moments about this point in each scenario. As a result of this, it is worth adding some extra distances onto our diagram, as shown. In both cases, we will let moments in the counterclockwise direction be positive. And since the reaction force 𝑅 is acting at the point at which we’re taking moments, this will have a moment equal to zero as its distance from the point 𝑑 is zero.

The 16-newton force acts in the counterclockwise direction about our point and will therefore have a positive moment. This is equal to 16 multiplied by the perpendicular distance 26 centimeters. The weight force acts in a clockwise direction about our point, so it will have a negative moment. This is equal to negative π‘Š multiplied by π‘₯ minus 26. The two-newton force will also have a negative moment and is equal to negative two multiplied by 51. The sum of these three moments equals zero. This equation simplifies to 416 minus π‘Šπ‘₯ plus 26π‘Š minus 102 is equal to zero. Adding π‘Šπ‘₯ to both sides and collecting like terms, we have π‘Šπ‘₯ is equal to 26π‘Š plus 314. We will call this equation one.

We will now repeat this process for the second scenario. The force at 𝐴 once again acts in the counterclockwise direction. So, we’ll have a positive moment equal to 18 multiplied by 23 centimeters. As already mentioned, the reaction force will have a moment equal to zero. So, the only other force we need to consider is the weight force. This acts in a clockwise direction about our point, giving us a moment of negative π‘Š multiplied by π‘₯ minus 23. Once again, the sum of our moments equals zero. And our equation simplifies to 414 minus π‘Šπ‘₯ plus 23π‘Š equals zero. Making π‘Šπ‘₯ the subject once again, we have π‘Šπ‘₯ is equal to 23π‘Š plus 414. We will call this equation two. And we now have a pair of simultaneous equations that we can solve to calculate the values of π‘Š and π‘₯.

As the right-hand side of our equations are equal, the left-hand sides must be equal. 26π‘Š plus 314 is equal to 23π‘Š plus 414. Subtracting 314 and 23π‘Š from both sides, we have three π‘Š is equal to 100. π‘Š is therefore equal to 100 over three. This is equal to 33.3 recurring, which to two decimal places is 33.33. The magnitude of the weight of the rod is 33.33 newtons. Rearranging equation one to make π‘₯ the subject, we have π‘₯ is equal to 26π‘Š plus 314 divided by π‘Š. We can now substitute our exact value of π‘Š into the right-hand side of this equation to calculate the value of π‘₯. This is equal to 35.42. The distance π‘₯ between the line of action of the weight and point 𝐴 is 35.42 centimeters. We now have the two required answers to this question rounded to two decimal places.

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