Question Video: Analysis of the Equilibrium of a Group of Couples Acting on a Rod Mathematics

𝐴𝐡 is a rod having a length of 90 cm and a negligible weight. It is suspended horizontally by a pin at its mid-point. Two forces each of magnitude 7.5 N, are acting at its ends. It is also pulled by a string, whose tension is 25 N in a direction making an angle of 30Β° with the rod from point 𝐢. If a force 𝐹 is acting on the rod at point 𝐷 so that the rod is in a horizontal equilibrium position, find the magnitude of 𝐹, its direction πœƒ, and the length of 𝐢𝐷.

09:44

Video Transcript

𝐴𝐡 is a rod having a length of 90 centimeters and a negligible weight. It is suspended horizontally by a pin at its midpoint. Two forces each of magnitude 7.5 newtons are acting at its ends, as shown in the figure. It is also pulled by a string whose tension is 25 newtons in a direction making an angle of 30 degrees with the rod from point 𝐢. If a force 𝐹 is acting on the rod at point 𝐷 so that the rod is in a horizontal equilibrium position, find the magnitude of 𝐹, its direction πœƒ, and the length 𝐢𝐷.

As the rod is in equilibrium, we know that the sum of the forces in the π‘₯- or horizontal direction equals zero. Likewise, the sum of the forces in the 𝑦- or vertical direction also equals zero. It will also be true that the sum of the moments about any point in the system will equal zero. We will take the positive directions to be to the right, vertically upward, and counterclockwise. We are told that the rod is of length 90 centimeters. If we let the point 𝑂 be the center or midpoint of the rod, we know that 𝐴𝑂 is equal to 𝑂𝐡, which is equal to 45 centimeters as this is a half of 90.

We will now clear some space so we can calculate the three unknowns required. The 25-newton force and 𝐹-newton force are not acting horizontally or vertically. This means that we need to find the horizontal and vertical components before we resolve in these directions. We can do this using our knowledge of right-angled trigonometry, where the sine of angle πœƒ is equal to the opposite over the hypotenuse and the cosine of angle πœƒ is equal to the adjacent over the hypotenuse.

If we consider the force at point 𝐢, we will let the horizontal component be equal to π‘₯ and the vertical component equal to 𝑦. This means that the sin of 30 degrees will be equal to 𝑦 over 25 and the cos of 30 degrees is equal to π‘₯ over 25. We can multiply both sides of these equations by 25. This means that 𝑦 is equal to 25 multiplied by the sin of 30 degrees, and π‘₯ is equal to 25 multiplied by the cos of 30 degrees. The sin of 30 degrees is equal to one-half, and the cos of 30 degrees is equal to root three over two. This means that we have a vertical component equal to 25 over two or 12.5 newtons and a horizontal component equal to 25 root three over two newtons.

We can repeat this for the 𝐹-newton force at point 𝐷. This time, we have a horizontal component acting to the left equal to 𝐹 cos πœƒ and a vertical component acting downward equal to 𝐹 sin πœƒ. We can now resolve in both the horizontal and vertical directions. The forces acting in a horizontal direction are 25 root three over two and 𝐹 cos πœƒ. As the sum of the forces is equal to zero and the 𝐹 cos πœƒ is moving to the left, we have 25 root three over two minus 𝐹 cos πœƒ is equal to zero. 25 over two is equal to 12.5, so this simplifies to 12.5 root three minus 𝐹 cos πœƒ is equal to zero. We can then add 𝐹 cos πœƒ to both sides, giving us an expression for this equal to 12.5 root three. We will call this equation one.

There are four forces acting in a vertical direction. From left to right, these are negative 7.5, positive 12.5, negative 𝐹 sin πœƒ, and positive 7.5. We know that these must also sum to equal zero. The two 7.5-newton forces are an example of a coplanar couple. They have the same magnitude but act in the opposite direction. This means that when resolving vertically, these forces will cancel as negative 7.5 plus 7.5 is equal to zero. Adding 𝐹 sin πœƒ to both sides of our equation gives us 12.5 is equal to 𝐹 sin πœƒ. We will call this equation two.

We now have a pair of simultaneous equations with two unknowns, 𝐹 and πœƒ. Dividing equation two by equation one, we get 12.5 over 12.5 root three is equal to 𝐹 sin πœƒ over 𝐹 cos πœƒ. We can divide the numerator and denominator of the left-hand side by 12.5, giving us one over root three. We can divide the numerator and denominator of the right-hand side by 𝐹. This leaves us with sin πœƒ over cos πœƒ, which we know is equal to tan πœƒ. We can then take the inverse tangent of both sides of this equation such that πœƒ is equal to the inverse tan of one over root three. This gives us a value of πœƒ equal to 30 degrees. We have calculated the direction of the force 𝐹.

Next, we need to substitute πœƒ equals 30 degrees into equation one or equation two. If we substitute into equation two, we have 12.5 is equal to 𝐹 sin 30. The sin of 30 degrees is equal to one-half. So, 12.5 is equal to 𝐹 multiplied by one-half. We can then divide both sides of this equation by one-half or multiply them both by two to give us a value of 𝐹 equal to 25 newtons. The magnitude of force 𝐹 is 25 newtons.

We will now clear some space and use moments to calculate the length of 𝐢𝐷. Before doing this, however, it is worth noting that the force 𝐹 and the 25-newton force are a coplanar couple. This is due to the fact they’re of equal magnitude and act in the opposite direction. The two forces are parallel to one another. This means that the system in this question consists of two pairs of coplanar couples. We will take moments about the midpoint 𝑂. But before doing this, we will tidy up our diagram so we have the relevant forces and distances.

As we have two pairs of coplanar couples, the distance 𝐢𝑂 must be equal to the distance 𝑂𝐷. We will call this distance π‘₯ centimeters. We know that the moment of a force is equal to its magnitude multiplied by the perpendicular distance. The force at 𝐴 is acting in a counterclockwise direction. Therefore, the moment is equal to 7.5 multiplied by 45. The force at 𝐢 is acting in a clockwise direction; therefore, this is equal to negative 12.5π‘₯. The vertical force at 𝐷 has the same magnitude. This is also acting in the clockwise direction and is equal to negative 12.5π‘₯. Finally, the moment of the force at 𝐡 is acting in a counterclockwise direction and once again is equal to 7.5 multiplied by 45. We know that the sum of these four values equals zero.

This simplifies to 337.5 minus 12.5π‘₯ minus 12.5π‘₯ plus 337.5 is equal to zero. Collecting like terms, we have 675 minus 25π‘₯ is equal to zero. Adding 25π‘₯ to both sides, we have 25π‘₯ is equal to 675. And then dividing both sides by 25 gives us π‘₯ is equal to 27. This means that the lengths of 𝐢𝑂 and 𝑂𝐷 are equal to 27 centimeters. 27 plus 27 is equal to 54. Therefore, the length 𝐢𝐷 is 54 centimeters. We now have our three answers. The magnitude of 𝐹 is 25 newtons, the direction πœƒ is 30 degrees, and 𝐢𝐷 is 54 centimeters.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.