In the figure below, 𝐹 sub one and 𝐹 sub two are two parallel forces measured in newtons, where 𝑅 is their resultant. If 𝑅 is equal to 30 newtons, 𝐴𝐵 equals 36 centimeters, and 𝐵𝐶 equals 24 centimeters, determine the magnitude of 𝐹 sub one and 𝐹 sub two.
In this question, we have two parallel coplanar forces, 𝐹 sub one and 𝐹 sub two, acting in opposite directions. We are also given the resultant force 𝑅, which is equal to 30 newtons. The distance from point 𝐴 to 𝐵 is 36 centimeters, and the distance from 𝐵 to 𝐶 is 24 centimeters. The lines of action of our forces 𝐹 sub one, 𝐹 sub two, and 𝑅 are not perpendicular to the line segment 𝐴𝐶. However, as our three forces are parallel, we can add the angle 𝜃 to our diagram as shown.
We can calculate the perpendicular components of these forces using our knowledge of right angle trigonometry. These are equal to 𝐹 sub one sin 𝜃, 𝐹 sub two sin 𝜃, and 𝑅 sin 𝜃. These will be useful when we come to take moments, as the moment of a force is equal to the magnitude of force multiplied by the perpendicular distance to the point at which we are taking moments. Whilst we can take moments about any point on our line, in this question, we will do so about point 𝐴.
We will consider moments acting in the counterclockwise direction to be positive and those acting in the clockwise direction to be negative. This means that the moment 𝑀 sub one of the force 𝐹 sub one acts in the positive direction and is equal to 𝐹 sub one sin 𝜃 multiplied by 36. We can repeat this process for 𝑀 sub two, which is the moment of the force 𝐹 sub two. As this acts in the negative direction, this is equal to negative 𝐹 sub two sin 𝜃 multiplied by 60.
Our expressions for 𝑀 sub one and 𝑀 sub two can be simplified as shown. We know that the distance 𝑥 from the line of action of the resultant force to the point at which we are taking moments is equal to the sum of the moments divided by 𝑅. In this case, we are taking moments about the point where the resultant acts. Therefore, 𝑥 is equal to zero. This means that the sum of our two moments must equal zero. 36 multiplied by 𝐹 sub one sin 𝜃 plus negative 60 multiplied by 𝐹 sub two sin 𝜃 equals zero. Since sin 𝜃 cannot be equal to zero, we can divide through by this. We can also divide through by 12 such that three 𝐹 sub one minus five 𝐹 sub two equals zero. As there are two unknowns here, we will call this equation one.
We will now consider the resultant force and the fact that this is equal to the sum of the other forces. Going back to our initial diagram, if we let the positive direction be vertically upwards, we have 𝑅 is equal to 𝐹 sub one plus negative 𝐹 sub two. Since 𝑅 is equal to 30 newtons, we have 30 is equal to 𝐹 sub one minus 𝐹 sub two. Adding 𝐹 sub two to both sides of this equation, we have 𝐹 sub one is equal to 30 plus 𝐹 sub two. We will call this equation two. And we now have a pair of simultaneous equations that we can solve by substitution.
One way of doing this is to substitute the expression for 𝐹 sub one in equation two into equation one. This gives us three multiplied by 30 plus 𝐹 sub two minus five 𝐹 sub two is equal to zero. Distributing the parentheses gives us 90 plus three 𝐹 sub two. The left-hand side then simplifies to 90 minus two 𝐹 sub two. By adding two 𝐹 sub two to both sides and then dividing through by two, we have 𝐹 sub two is equal to 45. Substituting this value back into equation two gives us 𝐹 sub one is equal to 30 plus 45, which is equal to 75. The magnitude of the two forces 𝐹 sub one and 𝐹 sub two are 75 newtons and 45 newtons, respectively.