Question Video: Finding Unknown Components of a Force Acting on a Body Moving Uniformly Where Two Other Force Are Acting on It | Nagwa Question Video: Finding Unknown Components of a Force Acting on a Body Moving Uniformly Where Two Other Force Are Acting on It | Nagwa

Question Video: Finding Unknown Components of a Force Acting on a Body Moving Uniformly Where Two Other Force Are Acting on It Mathematics • Third Year of Secondary School

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A body moves under the effect of two forces: 𝐅₁ = 8𝐒 + 𝐣 and 𝐅₂ = βˆ’4𝐒, where 𝐒 and 𝐣 are orthogonal unit vectors. A third force 𝐅₃ acts on the body, causing it to move uniformly. Find the angle 𝐅₃ makes with the unit vector 𝐒 rounding the result to the nearest minute.

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Video Transcript

A body moves under the effect of two forces 𝐅 sub one equals eight 𝐒 plus 𝐣 and 𝐅 sub two equals negative four 𝐒, where 𝐒 and 𝐣 are orthogonal unit vectors. A third force 𝐅 sub three acts on the body, causing it to move uniformly. Find the angle 𝐅 sub three makes with the unit vector 𝐒, rounding the result to the nearest minute.

Let’s begin by drawing a quick sketch representing what is happening to the body under the effect of these forces. We’re told that 𝐒 and 𝐣 are orthogonal unit vectors. We tend to model 𝐒 to act in the π‘₯-direction and 𝐣 to act in the 𝑦-direction. So the force 𝐅 sub one, which is eight 𝐒 plus 𝐣, might look a little something like this. Next, let’s consider our second force negative four 𝐒. Since 𝐒 is moving in the positive π‘₯-direction, negative four 𝐒 moves in the opposite direction as shown. So what about force 𝐅 sub three? We know that when this force acts on the body, the body moves uniformly. Now, it doesn’t really matter what happens before this third force acts. Just that when all three forces act on the body, it moves in a uniform manner.

Now, if the body is moving uniformly, we know that its acceleration is equal to zero. And in turn, the sum of the forces that act on the body must itself be equal to zero. So force 𝐅 sub three might look a little something like this. In order to calculate the angle it makes with the unit vector 𝐒, which we’ve modeled to be moving in the positive π‘₯-direction, we’re going to first find the sum of these forces and then set them equal to zero. That is, 𝐅 sub one plus 𝐅 sub two plus 𝐅 sub three equals zero. By substituting 𝐅 sub one with the force eight 𝐒 plus 𝐣 and 𝐅 sub two with negative four 𝐒, our vector equation becomes eight 𝐒 plus 𝐣 plus negative four 𝐒 plus 𝐅 sub three equals zero.

Now, in fact, we can add these two vectors by simply adding each individual component. Eight plus negative four is four, and the second force 𝐅 sub two has a 𝐣-component of zero. So the sum of those two vectors is four 𝐒 plus 𝐣. And when we add that to the force 𝐅 sub three, it’s equal to zero. To solve for 𝐅 sub three, we’ll subtract four 𝐒 plus 𝐣 from both sides. And once again, we’ll do this by subtracting the 𝐒-component and separately the 𝐣-component. That gives us force 𝐅 sub three is negative four 𝐒 minus 𝐣. In order to find the angle this force makes with the unit vector 𝐒, we’ll draw a second sketch. Let’s plot both vectors on the coordinate plane.

Remember, vector 𝐒 acts in the positive π‘₯-direction as shown. Then, the vector negative four 𝐒 minus 𝐣 acts four units to the left and one unit down. So in order to find the angle that 𝐅 sub three makes with the unit vector 𝐒, we’ll begin by finding the measure of this included angle in our right triangle. Let’s call that πœƒ. Then, we’ll be able to add 180 degrees to this angle since angles on a straight line sum to 180. Let’s clear some space and perform the next step.

Looking at the right triangle that we’ve drawn, we see that we know the measure of the opposite and adjacent sides. So we can use the tangent ratio to find the value of πœƒ. tan πœƒ is opposite over adjacent. So in this case, tan of πœƒ is one over four. It’s equal to a quarter. We can solve this equation for πœƒ by taking the inverse tan of both sides. And, of course, the inverse tan of tan πœƒ is simply equal to πœƒ. So πœƒ is equal to the inverse tan of one-quarter, which is 14.036 and so on.

Remember, to find the angle that 𝐅 sub three makes with the unit vector 𝐒, we add 180 degrees. Now, it’s worth noting that we do this because when we’re measuring angles on the coordinate plane, we tend to move in a counterclockwise direction from the positive π‘₯-axis. 180 plus 14.036 is 194.036. Now, we are going to need to convert this into the nearest minute. So we take the decimal part 0.036 and we times it by 60. That gives us 2.17 minutes. So correct to the nearest minute, that’s going to be equal to two. So the angle that force 𝐅 sub three makes with the unit vector 𝐒 correct to the nearest minute is 194 degrees and two minutes.

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