Question Video: Finding the Magnitude of Two Identical Forces given Their Resultant at Two Cases Mathematics

Two forces, both of magnitude 𝐅 N, act at the same point. The magnitude of their resultant is 90 N. When the direction of one of the forces is reversed, the magnitude of their resultant is 90 N. Determine the value of 𝐅.

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Video Transcript

Two forces, both of magnitude 𝐅 newtons, act at the same point. The magnitude of their resultant is 90 newtons. When the direction of one of the forces is reversed, the magnitude of their resultant is 90 newtons. Determine the value of 𝐅.

In this question, we have been told that there are two forces both with a magnitude of 𝐅 newtons such that their resultant is 90 newtons. Given that both of the forces act at the same point, we can sketch our two forces. We can label the angle between the forces as 𝜃 and add the resultant force of 90 newtons.

Recall that the law of cosines for two forces, 𝐅 sub one and 𝐅 sub two, where the angle between 𝐅 sub one and 𝐅 sub two is 𝛼 and the resultant of 𝐅 sub one and 𝐅 sub two is 𝐑 tells us that 𝐑 squared is equal to 𝐅 sub one squared plus 𝐅 sub two squared plus two 𝐅 sub one 𝐅 sub two cos 𝛼. In our case, we have that 𝐅 sub one and 𝐅 sub two are both equal to 𝐅. The angle between 𝐅 sub one and 𝐅 sub two, 𝛼, is equal to 𝜃. And the resultant force, 𝐑, is equal to 90.

Substituting these values into our formula, we have that 90 squared equals 𝐅 squared plus 𝐅 squared plus two 𝐅 squared cos 𝜃. Simplifying and factoring this equation, we have that 90 squared is equal to two 𝐅 squared lots of one plus cos 𝜃.

Now we can use the other information given to us in the question, which tells us that when the direction of one of the forces is reversed, the magnitude of the resultant is again 90 newtons. We have reversed the direction of the horizontal force. So now we can see that the angle between the two forces is 180 degrees minus 𝜃. Since this angle would have been the same if we reversed the direction of the other force, it does not matter which force we choose to reverse.

We again have that 𝐅 sub one and 𝐅 sub two are both equal to 𝐅 and 𝐑 is equal to 90. But 𝛼 is equal to 180 degrees minus 𝜃. Substituting these values into our formula, we have 90 squared equals 𝐅 squared plus 𝐅 squared plus two 𝐅 squared cos of 180 degrees minus 𝜃. We can simplify and factor this to obtain that 90 squared is equal to two 𝐅 squared lots of one plus cos of 180 degrees minus 𝜃. We can simplify this further using the fact that cos of 180 degrees minus 𝜃 is equal to negative cos of 𝜃. Our simplified equation becomes 90 squared equals two 𝐅 squared lots of one minus cos 𝜃.

We can move the two equations that we have found next to each other so that we can compare them. Looking at the two equations, the only difference is the sign in front of the cos 𝜃. Since both of these equations must be true, it must be the case that cos 𝜃 is equal to negative cos 𝜃. We add cos 𝜃 to both sides and divide by two to obtain cos 𝜃 equals zero. We know that 𝜃 will be in the range between zero degrees and 360 degrees. So solving cos 𝜃 equals zero, we have that 𝜃 is equal to 90 degrees or 270 degrees.

We can see that regardless of which angle we choose for 𝜃, the force diagram will look the same. Let us choose 𝜃 equals 90 degrees. We can update our force diagram to show this angle of 90 degrees. And we can substitute this value of 90 degrees into our original equation to find the value of 𝐅. We can simplify this by using the fact that cos of 90 degrees equals zero.

Next, we divide both sides by two. Next, we can take the square root of both sides. We can ignore the negative result, since 𝐅 is the magnitude of the force, so it must be positive. This square root can be simplified so that we have 𝐅 is equal to 90 over the square root of two. In order to rationalize the denominator of this fraction, we multiply the whole fraction by the square root of two over the square root of two. Finally, we simplify this to reach our solution, which is that 𝐅 equals 45 root two. There is, however, another way we could have solved this question.

At the point where we found our two equations in 𝐅 and 𝜃, if we add the two equations together, then the 𝜃 terms will cancel out. Let’s clear some space to explore this method.

We start by adding the two equations together. Next, we will factor the two 𝐅 squared terms on the right-hand side. We divide both sides of the equation by two and cancel out the cos 𝜃 terms. We are left with 90 squared is equal to two 𝐅 squared, which is the same equation we found when we were first solving the problem. We simplify this equation and take the square root of both sides. This rearranges to give us 45 root two, which matches the original solution we found.