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The Resultant of Two Forces - Problem 1
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
So one really important application of adding vectors, is finding the resultant of two forces in a Physics class. So let's find the resultant of the two forces g and f. Now f has a magnitude of 300 Newtons, g has a magnitude of 150 newtons and there's an angle of 110 degrees between them.
Finding the resultant means, finding the vector that these two add up to. We have to find both of its length and its direction. So let's recall the parallelogram method for finding the sum of two vectors.
What I'm going to do is, I'm going to duplicate each of these vectors on the other side and create a parallelogram. So I want a vector that's the same length and direction as f and I just need to approximate that. Then one that's the same length, and direction as g. This is the same direction and length as f, this is the same length and direction as g.
Now that I have my parallelogram, the sum of the two forces, f and g is going to start here at their common point of origin and finish here. So this is f plus g. So I need to find its length, and also its direction which I'll call this angle theta. I'll be able to say the direction of the force is so many degrees above horizontal.
Let's start with the length. Now I can find the length of this vector if I look at this triangle here. Now I know the length of this side, and I know the length of this side. It's the same as this one, 150 Newtons.
If I knew the measure of this angle, I would be able to find the length of the side opposite. Well, I do know the measure of that angle. This is a parallelogram. Two consecutive angles in a parallelogram have to be supplementary. They have to add up to 180. This is 110, so this has to be 70. That tells me that I can use the law of cosines to find the magnitude of f plus g. So the magnitude of f plus g squared which is 300² plus 150² minus twice the product 2 times 300 times 150 times the cosine of the angle between them; cosine 70.
So let me calculate that on my calculator. 300² plus 150² minus 2 times 300 times 150 times cosine 70. I get 81718 approximately. Now keep that value in your calculator, because now we have to take the square root. I get 285.9 Newtons. So the magnitude of f plus g, 285.9. Now keep that stored on your calculator as well, because we're going to need to find this angle. We'll use this measurement to find it.
Let's take a look at our picture again. I can actually now use the law of sines. Now that I know all three sides, I can use the law of sines to find this angle theta. Observe that sine of theta of this angle divided by the length of the opposite side, 150, equals the sine of this angle; 70 divided by the length of its opposite side. That's the magnitude of f plus g. Now I've got stored on my calculator, so I'll use that in a second.
First, let's multiply the 150 out. I get sine theta equals 150 sine 70 over the magnitude of f plus g. Then I've got to use inverse sine to get my theta. Theta is inverse sine of 150 sine 70 all over the magnitude of f plus g. Here goes.
So I've got inverse sine 150 times sine 70 divided by, and I'll use my stored value. I get 29.5 degrees. That's an approximation. 29.5 degrees. That means I'm done, because to describe the resultant, to find the resultant is to find both its length, its magnitude, 285.9 Newtons, and its direction. Its direction is 29.5 degrees above the horizontal.
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