𝐴𝐵 is a rod of negligible weight and length 54 centimeters. It is suspended horizontally by a pin at its midpoint. Forces of magnitude 68 root three newtons act on each end, one of them vertically upward at 𝐴 and the other vertically downward at 𝐵. The rod is pulled by a string, attached to it at point 𝐶, inclined at an angle of 60 degrees to 𝐴𝐵. The tension in the string has a magnitude of 192 newtons. The rod is kept in horizontal equilibrium by a fourth force 𝐹 acting on the rod at point 𝐷 with an angle of 60 degrees to 𝐵𝐴. Assuming that there is no reaction at the pin, find the magnitude of 𝐹 and the length of line segment 𝐷𝐶.
There is a lot of information in this question. So we will begin by drawing a free-body diagram. We are told that rod 𝐴𝐵 is 54 centimeters long and has negligible weight. It is suspended horizontally by a pin at its midpoint 𝑀. There are vertical forces at 𝐴 and 𝐵 of magnitude 68 root three newtons. The force at 𝐴 acts upwards and the force at 𝐵 acts downwards. The rod is pulled by a string attached to it at point 𝐶. The tension in the string is 192 newtons. And the string is inclined at an angle of 60 degrees to 𝐴𝐵. Finally, there is a fourth force 𝐹 acting on the rod at point 𝐷 with an angle of 60 degrees to 𝐵𝐴.
The forces at 𝐴 and 𝐵 are an example of a coplanar couple as they are equal in magnitude but opposite in direction. This means that when we resolve vertically, the forces will cancel. There are two further forces acting on the rod at points 𝐶 and 𝐷. These make the same angle with the rod and are therefore parallel. And we are also told the system is in equilibrium. This means that these forces must also form a coplanar couple. This means that they have the same magnitude and 𝐹 is equal to 192 newtons. We can check this by clearing some space and resolving vertically and horizontally. When a system is in equilibrium, we know that the sum of the forces in the horizontal direction equals zero and the sum of the forces in the vertical direction equals zero.
Before resolving vertically and horizontally, we need to use our knowledge of right angle trigonometry to find the horizontal and vertical components of the 192-newton force and the force 𝐹. The sin of angle 𝜃 is equal to the opposite over the hypotenuse, and the cos of angle 𝜃 is equal to the adjacent over the hypotenuse. If we consider the force 𝐹, we have a right triangle as shown. The horizontal component of this force is adjacent to the 60-degree angle. And the vertical component is opposite the 60-degree angle. This means that the sin of 60 degrees is equal to 𝑦 over 𝐹, and the cos of 60 degrees is equal to 𝑥 over 𝐹.
We know that the sin of 60 degrees is root three over two and the cos of 60 degrees is one-half. We can multiply both sides of our two equations by 𝐹 giving us 𝑦 is equal to root three over two 𝐹 and 𝑥 is equal to a half 𝐹. These are the vertical and horizontal components of force 𝐹, respectively. We can repeat this process for the 192-newton force. This has a horizontal component equal to 192 multiplied by the cos of 60 degrees, which is equal to 96 newtons acting to the left. The vertical component is equal to 192 multiplied by the sin of 60 degrees, which is equal to 96 root three newtons acting vertically upwards.
We can now resolve horizontally and vertically. If we let the positive direction act to the right, the sum of the horizontal forces is therefore equal to a half 𝐹 minus 96. And we know this is equal to zero. Adding 96 to both sides and then multiplying through by two, we get 𝐹 is equal to 192. This confirms our answer for 𝐹 of 192 newtons. We could also have resolved vertically. And taking the positive direction to be vertically upwards, we have the following equation. 68 root three minus root three over two 𝐹 plus 96 root three minus 68 root three equals zero. The 68 root threes cancel. We can then divide through by root three. This gives us negative 𝐹 over two plus 96 is equal to zero. Adding 𝐹 over two to both sides and then multiplying through by two once again gives us an answer of 192 newtons.
The second part of our question asks us to calculate the length of the line segment 𝐷𝐶. In order to do this, we will take moments about some point on the rod. As the rod is in equilibrium, we know that the sum of the moments equals zero. And we will take the positive direction to be counterclockwise. We are dealing with coplanar couples, so the forces of equal magnitude must be equidistant from the center of the rod. If we let the length of 𝐷𝑀 and 𝑀𝐶 be 𝑥 centimeters, then the length 𝐷𝐶 that we are trying to calculate is equal to two 𝑥 centimeters.
We know that the moment of any force is equal to the magnitude of the force multiplied by the perpendicular distance to the point at which we are taking moments. Whilst we can take moments about any point on the rod, in this case we will take moments about point 𝑀. The forces of magnitude 96 root three newtons are acting in a counterclockwise direction about this point. This means they will have a positive moment. In both cases, we have 96 root three multiplied by 𝑥. This is the same as two multiplied by 96 root three 𝑥, which in turn simplifies to 192 root three 𝑥.
The other two forces act in a clockwise direction about point 𝑀. This means they will have a negative moment. Both of these forces act at perpendicular distance of 27 centimeters from 𝑀. This means that they have a moment of negative 68 root three multiplied by 27. As the sum of the moments equals zero, we have 192 root three 𝑥 minus 3672 root three is equal to zero. We can divide through by root three and then add 3672 to both sides of our equation. Dividing through by 192 gives us 𝑥 is equal to 19.125. We can now substitute this in to calculate the length of 𝐷𝐶. 𝐷𝐶 is equal to 38.25 centimeters.
Using our knowledge of coplanar couples and by resolving and taking moments, we found that the magnitude of force 𝐹 is 192 newtons, and the length of line segment 𝐷𝐶 is 38.25 centimeters.