Question Video: Determining the Moment of a Couple Equivalent to a System of Forces Acting on a Triangle | Nagwa Question Video: Determining the Moment of a Couple Equivalent to a System of Forces Acting on a Triangle | Nagwa

Question Video: Determining the Moment of a Couple Equivalent to a System of Forces Acting on a Triangle Mathematics • Third Year of Secondary School

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π΄π΅πΆ is a triangle, where the side π΄π΅ = 20 cm, π΅πΆ = 25 cm, and πΆπ΄ = 15 cm. And forces of magnitudes 120, 150, and 90 newtons are acting along π΄π΅, π΅πΆ, and πΆπ΄, respectively, where the system is equivalent to a couple. Determine the magnitudes of the two parallel forces that would make the system in equilibrium when acting at π΅ and πΆ perpendicularly to π΅πΆ.

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Video Transcript

π΄π΅πΆ is a triangle, where the side π΄π΅ equals 20 centimeters, π΅πΆ equals 25 centimeters, and πΆπ΄ equals 15 centimeters. And forces of magnitudes 120, 150, and 90 newtons are acting along π΄π΅, π΅πΆ, and πΆπ΄, respectively, where the system is equivalent to a couple. Determine the magnitudes of the two parallel forces that would make the system in equilibrium when acting at π΅ and πΆ perpendicularly to π΅πΆ.

Letβs start with a diagram of the scenario. We have the triangle π΄π΅πΆ with side lengths π΄π΅ equals 20 centimeters, π΅πΆ equals 25 centimeters, and πΆπ΄ equals 15 centimeters. The force along π΄π΅ is 120 newtons. The force along π΅πΆ is 150 newtons. And the force along πΆπ΄ is 90 newtons. We have two parallel forces πΉ one and πΉ two, acting from the points π΅ and πΆ and perpendicular to π΅πΆ. The other forces acting on the triangle are equivalent to a couple. So in order to bring the system to equilibrium, the two forces acting at π΅ and πΆ must also be equivalent to a couple of equal magnitude and opposite direction. So πΉ one is equal to πΉ two. Letβs call this magnitude just πΉ.

To find the magnitude πΉ, we need to find the magnitude of the couple formed by the other forces acting on the triangle. To do this, we can calculate the moments of all three forces about a point of our choice and then add them together. Letβs calculate the moments of the three forces about the point π΄, taking the convention that counterclockwise moments are positive.

Recall that the magnitude of the moment π of a force πΉ acting from a point π is equal to the magnitude of the force πΉ multiplied by the perpendicular distance π between the pivot point π and the line of action of πΉ. The lines of action of the forces of 90 newtons and 120 newtons pass through the point π΄. So their perpendicular distances from π΄ are both zero. And hence, their moments about π΄ are also both zero. The only force that has a nonzero moment about π΄ is the force of 150 newtons. We need to find the perpendicular distance from the line of action of this force to the point π΄. This is equivalent to finding the altitude of the triangle with the line π΅πΆ as the base.

Recall that the altitude of a scalene triangle, that is, a triangle with three sides of different lengths, is given by two over π multiplied by the square root of π  times π  minus π times π  minus π times π  minus π, where π  is the semiperimeter of the triangle. π is the base of the triangle. And π and π are the other two sides of the triangle. The semiperimeter, π , is half the perimeter of the triangle, which is equal to 15 plus 20 plus 25 over two, which is equal to 30 centimeters. π is the length of the base line π΅πΆ, which is 25 centimeters. π and π are the lengths of the other two sides, which are 20 centimeters and 15 centimeters.

So, the altitude of the triangle, which is equivalent to the perpendicular distance π between the pivot point π΄ and the line of action of the force of 150 newtons, is given by two over 25 multiplied by the square root of 30 times 30 minus 20 times 30 minus 25 times 30 minus 15. This comes to the nice round number, 12 centimeters. The magnitude of the moment π is therefore given by the magnitude of the force, 150, multiplied by this perpendicular distance, 12, which is equal to 1800 newton-centimeters. Therefore, the magnitude of the resultant moment on the system is 1800 newton-centimeters. And the magnitude of the couple of the two parallel forces introduced at the points π΄ and π΅ must therefore also be 1800 newton-centimeters.

Recall that the magnitude of a couple acting at two points π΅ and πΆ with the forces at an angle π between their lines of action and the line π΅πΆ is equal to the magnitude of one of the forces, πΉ, multiplied by the length of the line π΅πΆ multiplied by its sin π. In this case, the two forces are perpendicular to the line π΅πΆ. So the angle π is equal to 90 degrees, and therefore sin π is equal to one. We can therefore simplify this to πΉ times the length of the line π΅πΆ. We can rearrange this equation for πΉ to give πΉ equals the magnitude of the couple, π π΅πΆ, over the length of the line π΅πΆ. This is equal to 1800 over 25. Therefore, the magnitudes of the two parallel forces that would make the system in equilibrium are both 72 newtons.

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