Question Video: Finding the Magnitude of a Resultant Force | Nagwa Question Video: Finding the Magnitude of a Resultant Force | Nagwa

# Question Video: Finding the Magnitude of a Resultant Force Physics • First Year of Secondary School

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The force π is the resultant of the two force vectors shown in the diagram. What is the magnitude of π to the nearest newton?

09:50

### Video Transcript

The force π is the resultant of the two force vectors shown in the diagram. What is the magnitude of π to the nearest newton?

In this question, weβve got a diagram that shows three vectors. This purple vector here represents a force π, and weβre told that π is the resultant of the other two force vectors. So thatβs the red arrows here, which are each labeled with a magnitude and a direction. Weβre being asked to find the magnitude of the vector π. The magnitude of π is the length of this purple arrow. We can see from the diagram that since π is the resultant of the other two vectors that when these two vectors are drawn tip to tail like this β so thatβs the second vector drawn, starting with its tail at the tip of the first vector β then the resultant vector π goes from the tail of the first vector to the tip of the second vector.

To find the magnitude of π or the length of this arrow, we can notice that the vector π forms the hypotenuse of a right-angled triangle. We can recall that Pythagorasβs theorem tells us that for a general right-angled triangle with a hypotenuse of length π and other sides of lengths π and π that π squared is equal to π squared plus π squared. If we take the square root of both sides of this expression, we have that π, the length of the hypotenuse, is equal to the square root of π squared plus π squared. In the case of the right-angled triangle that weβve identified in this diagram, the length of the hypotenuse is the magnitude of the vector π. So, if we can work out the lengths of the other two sides of the triangle, which weβve labeled as π and π, then Pythagorasβs theorem tells us that for this triangle the magnitude of π is equal to the square root of π squared plus π squared.

In order to find the values of π and π, weβre going to need to identify two more right-angled triangles in our diagram. The first of these, which weβve labeled as triangle one, is the right-angled triangle that has the first of the two red vectors as its hypotenuse. The other triangle, which weβve labeled as two, is the right-angled triangle whose hypotenuse is the second of the two red vectors. For each of these two triangles that weβve identified, weβre given the length of the hypotenuse and the value of one of the angles. Since this information isnβt particularly clear in the diagram now that weβve drawn over it, letβs draw out these two triangles separately from the diagram.

Triangle one looks like this kind of shape. And looking back at the diagram we were given, we can see that the leftmost angle in the triangle is 20 degrees, and the length of the hypotenuse, which is the magnitude of the first red vector, is 70 newtons. So, this is triangle one. Letβs now do the same thing for the triangle we labeled two. In this triangle, the leftmost angle is 70 degrees, and the length of the hypotenuse, which is the magnitude of the second red vector, is 60 newtons. So then, this is triangle two. If we now look back at the triangles in this diagram, we can see that π, the length of the horizontal side in the orange triangle, is equal to the sum of the horizontal side of triangle one plus the horizontal side of triangle two.

We could imagine shifting triangle two vertically downward so that the base or horizontal side of triangle two was at the same vertical height as the horizontal side of triangle one. And we would find that it slotted perfectly into this space here. Letβs label the horizontal side in triangle one as π one and in triangle two as π two. Then, in this diagram, we can identify the lengths π one and π two and then can easily see that π must be equal to π one plus π two. We can do the same thing for the vertical sides of the triangles one and two. Weβll label these as π one and π two, respectively. In the same way as before, in this diagram, we could imagine shifting triangle one horizontally until its vertical side lined up with the vertical side of triangle two. We can then identify the lengths π one and π two on this diagram. And we can see that π must be equal to π one plus π two.

Letβs quickly take a moment to take stock of the situation so far. Weβve got this equation here, which came from Pythagorasβs theorem. And it tells us how to calculate the magnitude of the force π if we know the values of the quantities weβve labeled as π and π. Then, weβve also got these two equations here, which tell us how to calculate π and π if we know the values of π one and π two and π one and π two. So, our approach is going to be to start by finding the values of π one, π two, π one, and π two. Then, weβll use these values to work out the quantities π and π. And then finally, we can use the values of π and π to work out the magnitude of the force π.

To find the values of these quantities π one, π two, π one, and π two, weβre going to need to use a couple of trigonometrical formulas. Letβs consider a general right-angled triangle, and weβll suppose that this angle has some value of π. Weβve labeled the side of the triangle opposite this angle as capital π for opposite, the side adjacent to the angle as capital π΄ for adjacent, and the hypotenuse as capital π» for hypotenuse. The two formulas that weβre going to need are these ones right here. The first one says that cos π is equal to π΄ divided by π». So, the cos of this angle π is equal to the length of the side thatβs adjacent to it divided by the length of the hypotenuse.

The second equation says that sin π is equal to π divided by π». So, the sin of this angle π is equal to the length of the side thatβs opposite it divided by the length of the hypotenuse. Looking at the triangles weβve labeled as one and two, we can see that the lengths weβre trying to find are the adjacent and opposite sides on these triangles. This means that we want to rearrange these two equations in order to make the quantities π΄ and π the subject. For both of these two equations, the process is the same. We simply need to multiply both sides by the hypotenuse π». When we do this rearrangement, we find that the adjacent side π΄ is equal to π» times cos π and the opposite side π is equal to π» times sin π.

Letβs now apply these two equations to our triangles one and two, starting with triangle one. In this case, the length of the hypotenuse is 70 newtons, and the angle which corresponds to π is 20 degrees. Then, the adjacent side, π one, must be equal to the hypotenuse of 70 newtons multiplied by the cos of the angle of 20 degrees. Similarly, the side opposite the angle, which is π one, must be equal to 70 newtons, the hypotenuse, multiplied by the sin of the angle 20 degrees. Evaluating the two expressions for the sides π one and π one, we find that π one is equal to 65.778 newtons and π one is equal to 23.941 newtons. In each case, the ellipses are used to indicate that the answers have further decimal places.

Letβs now clear some space and do the same thing for triangle two. In this case, the hypotenuse has a length of 60 newtons and the angle which corresponds to π has a value of 70 degrees. So, we have that the adjacent side π two is equal to 60 newtons multiplied by the cos of 70 degrees and the opposite side π two is equal to 60 newtons multiplied by the sin of 70 degrees. π two works out as 20.521 newtons, and π two works out as 56.382 newtons. Now that we found the values for the quantities π one, π two, π one, and π two, letβs clear ourselves some space so that we can use those values in these two equations to calculate the values of π and π.

We know that π is equal to π one plus π two. And substituting in these values for the quantities π one and π two and then evaluating the expression that we get, we find that π is equal to 86.299 newtons. In the same way, we know that π is equal to π one plus π two. And substituting in our values for π one and π two and then evaluating the expression, we find that π is equal to 80.323 newtons. Weβre going to clear ourselves some space one last time so that we can take our values for the quantities π and π and substitute them into this equation to calculate the magnitude of the force π.

Okay, so weβve got our value for π which is the horizontal side of the orange triangle. And weβve got our value for π, which is the vertical side of the same triangle. And weβve got our equation from Pythagorasβs theorem, which tells us that the hypotenuse of this triangle, which is the magnitude of the force π, is equal to the square root of π squared plus π squared. If we now substitute in our values for π and π, we get this expression for the magnitude of π.

Evaluating this sum under the square root, we find that the magnitude of π is equal to the square root of 13899.30 newtons squared. Taking the square root gives a result to two decimal places of 117.90 newtons. Weβre told in the question to give our answer to the nearest newton. By rounding this result to the nearest newton, we get our answer to the question that the magnitude of the force π is equal to 118 newtons.

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