Question Video: Finding the Magnitude of Two Forces in a Coplanar Couple | Nagwa Question Video: Finding the Magnitude of Two Forces in a Coplanar Couple | Nagwa

Question Video: Finding the Magnitude of Two Forces in a Coplanar Couple Mathematics • Third Year of Secondary School

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𝐴𝐡𝐢𝐷 is a rectangle, in which 𝐴𝐡 = 27 cm and 𝐡𝐢 = 18 cm. Forces with magnitudes 𝐹₁, 14 N, 𝐹₂, and 14 N are acting along 𝐴𝐡, 𝐡𝐢, 𝐢𝐷, and 𝐷𝐴 respectively. If this system of forces is in equilibrium, find 𝐹₁ and 𝐹₂.

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Video Transcript

𝐴𝐡𝐢𝐷 is a rectangle, in which 𝐴𝐡 is equal to 27 centimeters and 𝐡𝐢 is equal to 18 centimeters. Forces with magnitudes 𝐹 one, 14 newtons, 𝐹 two, and 14 newtons are acting along 𝐴𝐡, 𝐡𝐢, 𝐢𝐷, and 𝐷𝐴, respectively. If this system of forces is in equilibrium, find 𝐹 one and 𝐹 two.

We begin by sketching rectangle 𝐴𝐡𝐢𝐷, where side length 𝐴𝐡 is equal to 27 centimeters and side length 𝐡𝐢 is equal to 18 centimeters. We can convert these to the standard units of meters by dividing both values by 100. There are forces with magnitudes 𝐹 one, 14 newtons, 𝐹 two, and 14 newtons acting along the four sides of the rectangle as shown. The four forces form two couples. And we recall that a couple is a pair of parallel but not coincident forces of equal magnitudes and opposite directions. And since the forces 𝐹 one and 𝐹 two form a couple, they must be equal in magnitude.

We also know that since the system of forces is in equilibrium, the sum of the moments of the two couples must equal zero, where the moment of a couple is equal to the magnitude of the force multiplied by the perpendicular distance between the two forces’ lines of action. Convention dictates that moments acting in a counterclockwise direction are positive.

By letting the moment of the couple formed by the two 14-newton forces be π‘š one and the moment formed by the second couple be π‘š two, we have π‘š one plus π‘š two is equal to zero. Since π‘š one acts in the counterclockwise direction, it is equal to the force of 14 newtons multiplied by the perpendicular distance of 0.27 meters. We note that this moment would be measured in the standard units of newton-meters.

We have assumed that π‘š two also acts in a counterclockwise direction. And letting the force be 𝐹, we have 𝐹 multiplied by 0.18. As already mentioned, the sum of the two moments equals zero. So our equation is as shown. 14 multiplied by 0.27 is 3.78. Our equation simplifies to 3.78 plus 0.18𝐹 equals zero. We can subtract 3.78 from both sides of our equation. And dividing through by 0.18, we have 𝐹 is equal to negative 21.

As the force 𝐹 is equal to negative 21 newtons, we need to consider what this means. Our question specified that the two unknown forces were acting along 𝐴𝐡 and 𝐢𝐷. However, this does not necessarily mean that they have a positive magnitude in these directions. In actual fact, the positive directions for these two forces are 𝐡𝐴 and 𝐷𝐢. Looking at our diagram, this makes sense, since in order for the system to be in equilibrium, one of the couples must be acting in the counterclockwise direction and one in the clockwise direction. Since the forces we are trying to calculate, 𝐹 one and 𝐹 two, must be positive, these must both be equal to the magnitude of 𝐹. And since this is equal to 21 newtons, we have 𝐹 one equals 21 newtons and 𝐹 two equals 21 newtons.

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