Question Video: Determining the Magnitude of the Resultant of a Group of Forces by Resolving Them | Nagwa Question Video: Determining the Magnitude of the Resultant of a Group of Forces by Resolving Them | Nagwa

Question Video: Determining the Magnitude of the Resultant of a Group of Forces by Resolving Them Mathematics • Second Year of Secondary School

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Determine the magnitude of the resultant of the forces shown in the figure that are measured in newtons.

03:24

Video Transcript

Determine the magnitude of the resultant of the forces shown in the figure that are measured in newtons.

The resultant is the vector sum of all the forces in our diagram. So, let’s begin by listing the three forces we have. Force one has a force of 20 root two newtons, and it’s acting in the positive 𝑦-direction. We therefore say that its vector is 20 root two 𝐣. Similarly, our second force is 26 root two newtons acting in the negative 𝑥-direction. We therefore say it’s negative 26 root two 𝐢. But what about force three? We’re actually told the magnitude of this force. It’s 15 newtons, and it’s acting at an angle of 45 degrees with the positive 𝑥-axis. What we’re therefore going to do is to split it into its 𝐢- and 𝐣-components. And there are two ways we could do this. We could use right-angle trigonometry.

Alternatively, by recognizing that we’ve added a right-angle triangle with a second angle of 45 degrees, we see that we have an isosceles triangle. We’re therefore going to use the Pythagorean theorem to find the size of 𝑥. The Pythagorean theorem says that the square of the hypotenuse is equal to the sum of the square of the other two sides. In this case then, 15 squared is equal to 𝑥 squared plus 𝑥 squared, or two 𝑥 squared equals 225. We divide both sides of this equation by two and then find the square root of both sides. Now, usually, we would look to find both the positive and negative square root. In this case, though, we’re just looking to find the lengths of the sides of our triangle. We’ll consider their signs in a moment.

We can simplify our fraction a little by finding the square root of the numerator and the denominator. And we get 𝑥 equals 15 over root two. By multiplying both the numerator and the denominator by root two, we end up with a rational denominator. And we find 𝑥 to be equal to 15 root two over two. Now, we can see that both the 𝐢- and 𝐣-components are acting in their respective positive directions. So, we can say that the vector for force three is 15 root two over two 𝐢 plus 15 root two over two 𝐣. Remember, the resultant is the vector sum of these forces. So, let’s call that resultant 𝑅. And we have 20 root two 𝐣 plus negative 26 root two 𝐢 plus 15 root two over two 𝐢 plus 15 root two over two 𝐣.

Our next job is to collect together the components for 𝐢 and 𝐣. We have 15 root two over two and negative 26 root two 𝐢, then 20 root two plus 15 root two over two 𝐣. And this simplifies giving us a resultant of negative 37 root two over two 𝐢 plus 55 root two over two 𝐣. We’re not quite finished, though. Remember, we’re looking to find the magnitude of this resultant. Essentially, the magnitude of a vector is its length. And we can therefore use the Pythagorean theorem to calculate this.

The magnitude of our resultant is the square root of the sum of the squares of each component, so negative 37 root two over two all squared plus 55 root two over two all squared. That’s the square root of 2197, which is 13 root 13. Remember, all the forces in this question are measured in newtons, so, so is the magnitude of the resultant. It’s 13 root 13 newtons.

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