Question Video: Analysis of the Equilibrium of a System of Forces acting on a Regular Hexagon Mathematics

𝐴𝐡𝐢𝐷𝐸𝐹 is a regular hexagon having a side length of 18 cm. Forces of magnitudes 18, 7, 18, and 7 newtons are acting along 𝐡𝐴, 𝐡𝐢, 𝐸𝐷, and 𝐸𝐹 respectively. Two other forces, each of magnitude 𝑃 newtons, are acting along 𝐢𝐷 and 𝐹𝐴. Find the value of 𝑃, given that the system is in equilibrium.

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Video Transcript

𝐴𝐡𝐢𝐷𝐸𝐹 is a regular hexagon having a side length of 18 centimeters. Forces of magnitudes 18, seven, 18, and seven newtons are acting along 𝐡𝐴, 𝐡𝐢, 𝐸𝐷, and 𝐸𝐹, respectively. Two other forces, each of magnitude 𝑃 newtons, are acting along 𝐢𝐷 and 𝐹𝐴. Find the value of 𝑃, given that the system is in equilibrium.

The diagram shows two couples of magnitudes 18 and seven newtons acting on the system. We know this since a couple is a pair of parallel but not coincident forces of equal magnitudes and opposite directions. We have 18-newton forces acting in parallel but opposite directions along 𝐡𝐴 and 𝐸𝐷. Likewise, there are seven-newton forces acting along 𝐸𝐹 and 𝐡𝐢. We are also told in the question that there is a third couple acting along 𝐢𝐷 and 𝐹𝐴. This has magnitude of 𝑃 newtons.

Since the system is in equilibrium, we know that the counterclockwise moment due to the couples must be equal to the clockwise moment due to the couples. Another way of saying this is that the sum of the moments equals zero. We know that the moment of a couple is equal to the magnitude of the force of the couple multiplied by the perpendicular distance between the lines of action. Since the hexagon is regular, it can be split into six equilateral triangles as shown. This means that the distance between point 𝐢 and point 𝐹, where the forces of 𝑃 newtons act, is 36 centimeters. Likewise, we have a distance of 36 centimeters between point 𝐡 and point 𝐸, where the 18-newton forces and seven-newton forces act.

The components of the forces acting perpendicular to these lines are as shown. Using our knowledge of trigonometry, these are equal to seven multiplied by sin of 60 degrees, 18 multiplied by sin of 60 degrees, and 𝑃 multiplied by sin of 60 degrees. We are now in a position to calculate the moment of each couple, noting that counterclockwise moments are taken as positive. The 18-newton force couple acts in this direction, giving us a moment of 18 sin 60 degrees multiplied by 36. The seven-newton force couple acts in the clockwise direction and, therefore, has a negative moment. This is equal to negative seven sin 60 multiplied by 36. Finally, the 𝑃-newton force couple also acts in the clockwise direction and, therefore, has a moment equal to negative 𝑃 sin 60 multiplied by 36.

As already mentioned, the sum of these moments equals zero. And since all three expressions contain the sin of 60 degrees and 36, we can divide through by both of these. This gives us 18 plus negative seven plus negative 𝑃 is equal to zero. 18 plus negative seven is equal to 11. And adding 𝑃 to both sides, we have 𝑃 equals 11. Given that the system is in equilibrium, the value of 𝑃 is therefore equal to 11 newtons.

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