# Question Video: Finding the Couple That Puts a Rod in Equilibrium Mathematics

𝐴𝐵 is a rod having a length of 50 cm and a negligible weight. Two coplanar pairs of forces are acting on the rod as shown in the figure. The first couple consists of two forces acting perpendicularly to the rod, each of magnitude 2 kg-wt, and the second couple consists of two forces, each of magnitude 𝐹. Determine the value of 𝐹 that makes the rod in equilibrium.

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### Video Transcript

𝐴𝐵 is a rod having a length of 50 centimeters and a negligible weight. Two coplanar pairs of forces are acting on the rod as shown in the figure. The first couple consists of two forces acting perpendicularly to the rod, each of magnitude two kilogram-weight. And the second couple consists of two forces, each of magnitude 𝐹. Determine the value of 𝐹 that makes the rod in equilibrium.

We recall that a coplanar couple is two forces that are equal in magnitude and opposite in direction. In this question, we have two such couples. In order to calculate 𝐹, we’ll need to take moments about a point on the rod. We know that when the rod is in equilibrium, the sum of our moments is equal to zero. We know that a moment is equal to the force multiplied by the perpendicular distance. This means that we will firstly need to calculate the vertical components of the forces 𝐹.

Using our knowledge of right-angle trigonometry, we know that the sine of angle 𝜃 is equal to the opposite over the hypotenuse. This means that, in this question, the sin of 45 degrees is equal to 𝑦 over 𝐹. The vertical components are therefore equal to 𝐹 multiplied by sin of 45 degrees. The sin of 45 degrees is equal to root two over two. We will now take moments about point 𝐵, taking the counterclockwise direction to be positive.

The first force of magnitude two newtons will have a moment equal to two multiplied by 10 as it is 10 centimeters from 𝐵. Next, we have root two over two 𝐹 multiplied by 30. Our final force is acting in a clockwise direction, giving us negative two multiplied by 50. The sum of these forces equals zero. Our equation simplifies to 20 plus 15 root two 𝐹 minus 100 equals zero. 20 minus 100 is equal to negative 80, and we can add this to both sides. Dividing both sides by 15 root two gives us 𝐹 is equal to 16 over three root two. We can then rationalize the denominator, giving us a value of 𝐹 equal to eight root two over three kilogram-weight.