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Question Video: Studying the Equilibrium of a Horizontal Rod Resting by the Means of a Support and a String Mathematics

A uniform rod 𝐴𝐡 having a length of 62 cm and weighing 13 N is resting horizontally by means of a support and a string. Given that the support is at the end 𝐴 and the string is 8 cm away from the end 𝐡, determine the string’s tension 𝑇 and the support’s reaction 𝑅, rounding your answer to two decimal places.

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Video Transcript

A uniform rod 𝐴𝐡 having a length of 62 centimeters and weighing 13 newtons is resting horizontally by means of a support and a string. Given that the support is at the end 𝐴 and the string is eight centimeters away from the end 𝐡, determine the string’s tension 𝑇 and the support’s reaction 𝑅, rounding your answer to two decimal places.

We will begin by sketching the rod 𝐴𝐡, which has length 62 centimeters. As the rod is uniform, the weight acts at the midpoint of the rod. As the weight of rod 𝐴𝐡 is 13 newtons, there will be a downward force equal to this, 31 centimeters from 𝐴. We are told there is a support at end 𝐴. This will have a normal reaction force 𝑅 acting vertically upwards. The rod is also supported by a string with tension 𝑇, which is eight centimeters away from the end 𝐡.

Since 62 minus eight is 54, the string is 54 centimeters from end 𝐴. We have been asked to calculate the string’s tension 𝑇 and the support’s reaction 𝑅. Since the rod is resting horizontally, we know it is in equilibrium. And when any body is in equilibrium, the sum of the forces acting on it must equal zero. The sum of the moments must also equal zero. And the moment of any force is equal to 𝐹 multiplied by 𝑑, where 𝐹 is the force acting at a point and 𝑑 is the perpendicular distance from this point to the point at which we are taking moments.

Let’s begin by considering the sum of the forces in this question. If we let the positive direction be vertically upwards, we see that the reaction force 𝑅 and tension force 𝑇 are acting in this direction. The 13-newton weight force is acting vertically downwards, so this will be negative. Since the sum of the forces equals zero, we have 𝑅 plus 𝑇 minus 13 equals zero. By adding 13 and subtracting 𝑇 from both sides, we have 𝑅 is equal to 13 minus 𝑇. As there are two unknowns, we cannot go any further. And we will call this equation one.

Let’s now consider moments. Whilst we could take moments about any point, we will take moments about point 𝐴 and consider the positive direction to be counterclockwise. As the reaction force acts at point 𝐴, it will have a moment equal to zero, as the distance 𝑑 is zero. The 13-newton force acts in a clockwise direction about point 𝐴. This means it will have a negative moment equal to negative 13 multiplied by 31, as this force acts a perpendicular distance of 31 centimeters from 𝐴.

The tension force acts in a counterclockwise direction about point 𝐴. Therefore, it will have a positive moment. This is equal to 𝑇 multiplied by 54. Once again, we know that the sum of our moments equals zero. Our equation simplifies to negative 403 plus 54𝑇 equals zero. We can then add 403 to both sides of the equation. Dividing through by 54, we have 𝑇 is equal to 403 over 54, which is equal to 7.4629 and so on. To two decimal places, the tension force 𝑇 is equal to 7.46 newtons.

We can now substitute this value into equation one. Using the exact value for 𝑇 and subtracting it from 13 gives us 𝑅 is equal to 5.5370 and so on. To two decimal places, this rounds to 5.54. The reaction force 𝑅 is therefore equal to 5.54 newtons. The string’s tension 𝑇 is 7.46 newtons. And the support’s reaction 𝑅 is 5.54 newtons.

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