Question Video: Finding Missing Forces in a System Given that it is Equivalent to a Couple | Nagwa Question Video: Finding Missing Forces in a System Given that it is Equivalent to a Couple | Nagwa

# Question Video: Finding Missing Forces in a System Given that it is Equivalent to a Couple Mathematics • Third Year of Secondary School

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In the figure, π΄π΅πΆπ· is a right-angled trapezoid at π΄, where π΄π΅ = 12 cm, π΅πΆ = 32 cm, and π΄π· = 16 cm. The forces are measured in newtons and are represented completely by the sides of the trapezoid, where the magnitude of the forces is proportional to their corresponding side lengths. If the system of forces is equivalent to a couple, find πΉβ, πΉβ, and πΉβ.

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### Video Transcript

In the figure shown, π΄π΅πΆπ· is a right-angled trapezoid at π΄, where π΄π΅ equals 12 centimeters, π΅πΆ equals 32 centimeters, and π΄π· equals 16 centimeters. The shown forces are measured in newtons and are represented completely by the sides of the trapezoid, where the magnitude of the forces is proportional to their corresponding side lengths. If the system of forces is equivalent to a couple, find πΉ one, πΉ two, and πΉ three.

All right, so here we see this trapezoid with corners π΄, π΅, πΆ, and π·. And we also see the three unknown forces πΉ one, πΉ two, and πΉ three we want to solve for. Each of these forces acts along a side of the trapezoid along with this known force of 30 newtons. Our problem statement tells us that the side length π΄π΅ is 12 centimeters, π΅πΆ is 32 centimeters, and side length π΄π· is 16. Weβre told further that the magnitude of the forces acting on this trapezoid is proportional to their corresponding side lengths.

This means, for example, that since side length π΅πΆ is twice as long as side length π΄π·, then the magnitude of the force πΉ three must be twice as great as the magnitude of πΉ one. This same proportionality relationship applies to the other sides. And weβre also told that this system of four forces is equivalent to a couple. That means we can equate these four forces to two equal and opposite forces that act along different lines of action. Knowing all this, itβs the unknown forces πΉ one, πΉ two, and πΉ three we want to solve for.

To begin doing that, letβs clear some space on screen and letβs think about what it means that these four forces are equivalent to a couple. If we draw the lines of action of these four forces, they effectively trace out the sides of this trapezoid. Because these four forces are equivalent to a couple, that means we can pick two points of intersection of these lines of action. And so long as those points lie along all four lines of action, then we can say that a force couple effectively originates from those points.

Hereβs what we mean by that. Say we picked the corners of the trapezoid π΅ and π·. At point π΅, the lines of action of forces πΉ two and πΉ three meet. And point π·, we see, is an intersection of the lines of action of our 30-newton force and πΉ one. In that sense then, weβve accounted for all four forces. And so if we sketch in these forces as though they originate at these two corners of our trapezoid, they would look something like this.

So weβre modeling all four forces as though they start at these two corners of our trapezoid. Note that we could just as well have picked the corners π΄ and πΆ since those two corners intersect all four lines of action. But either way, we can analyze these four forces as a couple. All this means that the net force acting at point π· is equal in magnitude but opposite in direction to the net force acting at point π΅. And actually since forces in what we could call the vertical direction are independent of forces in the horizontal direction, we can also say that the total vertical force at π· is equal and opposite that at π΅, and likewise for the horizontal components.

This means there are two force balance equations we can write out, one for the vertical direction and one for the horizontal. If we decide that forces to the right and forces pointing upward are positive, then when it comes to the vertical forces acting at points π΅ and π·, we can write that the vertical component of our 30-newton force, weβll call it 30 sub v, minus the force πΉ two acting at point π΅ equals zero. Rearranging this equation slightly, we see then that if we can solve for that vertical component of our 30-newton force, weβll know πΉ two.

Coming back to our sketch, we see that this 30-newton force is effectively the hypotenuse of a right triangle. If we think of this triangle though not in terms of forces but in terms of distances, we know that this side is equal to 12 centimeters, this side is 32 minus 16 or 16 centimeters. And that means the hypotenuse is equal to the square root of 12 squared plus 16 squared, or 20 centimeters. Knowing these side lengths is helpful because if we call this interior angle of our right triangle π, then we can say that the vertical component of our 30-newton force is equal to 30 times the sin of π.

We recall that given a right triangle where another interior angle is π, the sin of that angle is equal to the ratio of the opposite side length to the hypotenuse length. This means that in our right triangle of interest, the sin of π is equal to the ratio of 12 divided by 20. 30 times 12 over 20 equals 18. And so we now know the magnitude of force πΉ two. And we know that this is in units of newtons. Weβll record this result and now move on to solving for the two remaining unknown forces πΉ one and πΉ three. Both of these forces, we see, are directed horizontally.

For our force balance equation, we can write that πΉ three minus πΉ one minus the horizontal component of the 30-newton force, 30 times the cos of π, all equal zero. Coming back to our example right triangle, we can say that the cos of this angle π is equal to the adjacent side length divided by the hypotenuse. In the triangle formed from our trapezoid, this ratio is equal to 16 divided by 20. Even knowing this though, we can see that here we have one equation but two unknowns. Weβll need to bring in some other information then to solve for πΉ one and πΉ three.

At this point, we can recall that the forces involved in this scenario are proportional to one another according to their corresponding side lengths. This means, for example, that the ratio of side length π΄π·, 16 centimeters, to the side length π΄π΅ is equal to the ratio of πΉ one to πΉ two, the forces on those respective sides. This is great because it means that πΉ one equals 16 over 12 times πΉ two, which we know. So πΉ one then equals 16 over 12 times 18 newtons, or 24 newtons.

Recalling this proportionality method then, there are now two ways we could go about solving for πΉ three. We could use the same method we used to solve for πΉ one. Or, now that we know πΉ one, we could use this force balance equation here. Since weβve already written this equation out, letβs continue on with it. Adding πΉ one and 30 times 16 over 20 to both sides, we get that πΉ three equals πΉ one plus 30 times 16 over 20. And knowing that πΉ one without units is 24 and that 16 over 20 equals four-fifths, we find that πΉ three is equal to 48 with units of newtons.

We find then that the three forces πΉ one, πΉ two, and πΉ three acting on the sides of this trapezoid have magnitudes of 24, 18, and 48 newtons, respectively. These magnitudes mean that all four forces involved with the trapezoid act as a couple.

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