##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- FREE study tips and eBooks on various topics

# The Resultant of Two Forces - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We're finding the resultant of two forces. Here I have an example; find the resultant of the two forces f which has a magnitude of 600 newtons, g which has a magnitude of 400 newtons and the angle between the two forces is 40 degrees. So what's the resultant?

Well, finding the resultant means finding the sum of the two vectors. So the first thing I want to do is set up a parallelogram. I'm going to use the parallelogram method here. So let me draw a vector that is parallel to f, and the same length as f. Then I'm basically drawing a copy of two vectors, I'll draw a copy of g and that gives me a parallelogram.

Now the sum of the two vectors, is going to be the vector that starts at the common initial point of the two vectors, and ends at the opposite side of the parallelogram. So this is the sum. This is the sum; f plus g. So to find the resultant means, to find the magnitude of this sum and the direction. Now I'll describe the direction by finding this angle here, the angle theta. I'll say the direction is so many degrees below the horizontal.

First, what's the magnitude? Well, let's observe that. We know the length of this side, and since it's a parallelogram, we know the length of this side. Parallelograms have opposite sides congruent. So this is going to be 400 Newtons and this side is going to be 600 Newtons. But we really only need to focus on this triangle right here. Now if I knew the measure of angle, I could find the length of this side using the law of cosines. But I do know the measure of this angle because this angle is the supplement of this angle, the 40 degree angle. So this one is going to be 140 degrees because the two have to add up to 180.

That means that by the law of cosines, the magnitude of plus g² is 600² plus 400², minus twice the product 2 times 600 times 400, times cosine of 140. So let me get an approximation for that on my calculator. 600² plus 400² minus 2 times 600 times 400, times cosine 140. I get 887,701, and then take the square root, I get 942. So this was 887,701 and the magnitude of f plus g is approximately 942.2 Newtons. I want to keep that value stored on my calculator because I'm going to need it in a second. So I'll store this as say w.

Now I still need to find theta which will give me the direction of my resultant force. So how will I find theta? Well I still have this triangle to work with, and now I know all three sides. So what I'm going too do is I'm going to use the law of sines to find theta.

Sine of theta, divided by the length of the side opposite, 400, equals the sine of 140, divided by the length of the side opposite that. And of course that's the value we just found, the length of f plus g.

So let me just multiply both sides by 400 and I get sine of theta equals 400 sine 140, over the length of f plus g, and then inverse sine will get me my answer. Theta is inverse sign of 400 sine 140 over the magnitude of f plus g, the value that I stored as w. So let's calculate this. Inverse sine 400 times sine 140 divided by w, I get 15.84 degrees. So this is an approximation 15.84 degrees. And that means that my resultant acts 15.84 degrees below the horizontal.

So my final description is, that it has a magnitude of 942.2 Newtons and it acts in a direction 15.84 degrees below the horizontal.

Please enter your name.

Are you sure you want to delete this comment?

###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

##### Concept (1)

##### Sample Problems (3)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete