Question Video: Finding the Tension in a String Holding a Rigid Body in Equilibrium Mathematics

𝐴𝐡 is a uniform rod of length 140 cm and weight 45 kg-wt. Its end 𝐴 is fixed to a vertical wall by a hinge. It is held horizontally in equilibrium by means of a string of length 70 cm connected to point 𝐢 on the rod which is 56 cm away from 𝐴, and fixed to point 𝐷 on the vertical wall vertically above 𝐴. Calculate the tension in the string.

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Video Transcript

𝐴𝐡 is a uniform rod of length 140 centimeters and weight 45 kilogram weight. Its end 𝐴 is fixed to a vertical wall by a hinge. It is held horizontally in equilibrium by means of a string of length 70 centimeters connected to point 𝐢 on the rod, which is 56 centimeters away from 𝐴, and fixed to point 𝐷 on the vertical wall vertically above 𝐴. Calculate the tension in the string.

To answer this question, we’re going to begin by drawing a free-body diagram. This is a really simple diagram of the scenario that shows all the important forces we’re interested in. Here is the uniform rod of length 140 centimeters attached to a vertical wall at point 𝐴 by a hinge. Now, since the rod itself is uniform, this means we can model the downwards force of its weight to act at a point exactly halfway along the rod. So this 45-kilogram-weight force acts at a point 70 centimeters away from 𝐴.

Next, we know that the rod itself is attached by a piece of string at point 𝐢 to a point on the wall 𝐷. The distance between 𝐴 and 𝐢 is 56 centimeters. And the distance between 𝐷 and 𝐢, the length of the string, is 70 centimeters. We’re being asked to calculate the value of the tension in the string. So we model that acting at point 𝐢, away from 𝐢, and along the length of the string, and we’ll call that 𝑇.

So now we have all the relevant forces and dimensions, how are we going to calculate the value of 𝑇? Well, the key here is this word equilibrium. For the object to be in equilibrium, the sum of its forces must be equal to zero. But also the sum of the moments about any point must also be equal to zero. This means we can find the value of 𝑇 by finding the sum of the moments about some point on the rod. Now, we’re going to take moments about point 𝐴, and we’re going to take the counterclockwise direction to be positive.

We also know, of course, that the moment is the product of a force and its perpendicular distance from the pivot. So, thinking about the weight force, we know that this moment is going to be negative because it’s trying to rotate the object in a clockwise direction. It’s a force of 45 kilogram weight, and it’s acting at a point 70 centimeters away from 𝐴. So its moment in kilogram weight centimeters is negative 45 times 70.

But what do we do with our tensional force? Well, we need to work out the component of this force that’s perpendicular to the line segment 𝐴𝐡. So we’re going to begin by calculating the value of πœƒ. That’s the included angle in triangle 𝐴𝐢𝐷. We can use the dimensions given to calculate this value. We know, of course, the side adjacent to our included angle and the hypotenuse. So we use the cosine ratio. cos of πœƒ is 56 over 70. We could then take the inverse cos of both sides of this equation. But in fact we can find some exact values by also calculating the length of the other side.

We’re dealing with a right triangle, so we’re going to use the Pythagorean theorem. Defining the missing length in this triangle to be π‘₯, 70 squared equals 56 squared plus π‘₯ squared. Subtracting 56 squared from both sides, and we get π‘₯ squared equals 1764. And the square root of this is 42, so π‘₯ is equal to 42. And this is really useful because it allows us to calculate exact values for the sin, cos, and tan of πœƒ. For instance, sin of πœƒ is the opposite side divided by the hypotenuse, so it would be 42 over 70. And tan is opposite over adjacent, so tan of πœƒ here is 42 over 56.

We’re now ready to calculate the component of the tensional force that acts in a direction perpendicular to the rod. It’s going to be 𝑇 sin πœƒ, since we have the hypotenuse 𝑇 and we’re trying to find the component opposite to the included angle. Since the tensional force is trying to rotate object 𝐴𝐡 in a counterclockwise direction, the moment of this tensional force is going to be positive. It’s 𝑇 sin πœƒ times 56. And now we have the sum of all of the moments, and this is equal to zero. Negative 45 times 70 is negative 3150. And remember, we calculated sin of πœƒ to be 42 over 70. So 𝑇 sin πœƒ times 56 becomes 𝑇 times 42 over 70 times 56. 42 over 70 times 56 simplifies to 168 over five. Then, we can add 3150 to both sides of our equation.

We clear some space and we see that we can calculate the value of 𝑇 by simply dividing through by 168 over five, giving us 93.75. So the tension in the string is 93.75 kilogram weight.

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