# Video: Analysis of a System of Four Forces Acting on a Horizontal Rod Equivalent to a Couple

𝐴𝐵 is a horizontal light rod having a length of 60 cm, where two forces, each of magnitude 45 N, are acting vertically at 𝐴 and 𝐵 in two opposite directions. Two other forces, each of magnitude 120 N, are acting in two opposite directions at points 𝐶 and 𝐷 of the rod, where 𝐶𝐷 = 45 cm. If they form a couple equivalent to the couple formed by the first two forces, find the measure of the angle of inclination that the second two forces make with the rod.

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### Video Transcript

𝐴𝐵 is a horizontal light rod having a length of 60 centimeters, where two forces, each of magnitude 45 newtons, are acting vertically at 𝐴 and 𝐵 in two opposite directions. Two other forces, each of magnitude 120 newtons, are acting in two opposite directions at points 𝐶 and 𝐷 of the rod, where 𝐶𝐷 equals 45 centimeters. If they form a couple equivalent to the couple formed by the first two forces, find the measure of the angle of inclination that the second two forces make with the rod.

Looking at our diagram, we’re told that the forces acting on points 𝐴 and 𝐵, which we can call 𝐹 sub one and negative 𝐹 sub one, have a magnitude of 45 newtons and form a force couple. Similarly, the forces acting at points 𝐶 and 𝐷, which we can call 𝐹 sub two and negative 𝐹 sub two, have a magnitude of 120 newtons and also form a force couple.

If we call the moment of the first couple 𝑀 sub one and the moment of the second couple 𝑀 sub two, we’re told that these two couples are equivalent, or their magnitudes are equal to one another. Based on this information, we want to find the measure of the angle of inclination that the second two forces make with the rod.

In our diagram, we’ve called this angle 𝜃. To solve for 𝜃, we can write out the first moment 𝑀 sub one and the second moment 𝑀 sub two and then set them equal to one another. Since 𝐹 one acts perpendicular to the horizontal rod, we can write that 𝑀 sub one equals the magnitude of 𝐹 sub one multiplied by the length of segment 𝐴𝐵.

We’re told that that segment is 60 centimeters long. Therefore, 𝑀 sub one is equal to 45 newtons multiplied by 60 centimeters. The form of the equation for 𝑀 sub two is similar, with the magnitude now of 𝐹 sub two and the length of line segment 𝐶 sub 𝐷 [𝐶𝐷]. But there’s an additional term we’ve added in.

This sin of 𝜃 term guarantees that the component of 𝐹 two we’re considering in this moment is perpendicular to the distance separating the lines of action. We’re told in the exercise statement that the distance 𝐶 to 𝐷 is 45 centimeters. So 𝑀 sub two equals 120 newtons times 45 centimeters times the sin of 𝜃.

Since the moments 𝑀 sub one and 𝑀 sub two are equivalent, we can set 45 newtons times 60 centimeters equal to 120 newtons times 45 centimeters times the sine of the angle we want to solve for, 𝜃.

Dividing both sides of the equation by 120 newtons times 45 centimeters and then taking the inverse sine of both sides, we find that 𝜃 is equal to the arcsin of 45 newtons times 60 centimeters over 120 newtons times 45 centimeters.

When we enter that value on our calculator, we find that 𝜃 is 30 degrees. That’s the angle our second pair of forces makes with the line segment 𝐶𝐷.