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# The Resultant of Two Forces - Problem 3

###### 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.

I'm finding the resultant of two forces. Here I have a problem that asks me to find the resultant of the forces; f and g which are holding up a street light. Both of these forces have magnitudes of 500 pounds, and both forces make an angle of 75 degrees with the vertical. So I have to find the sum of these two forces.

I have two choices, I can use the parallelogram rule, or I can put the vectors head to tail. This time, I think I'm going to put the vectors head to tail.

So I'm going to take a copy of g and out it up here. It doesn't have to be perfect, not a perfect drawing just reasonably accurate. So there is my copy of g, this is 500 pounds. Now the resultant of these two forces is going to start at the initial point of f, and end at the terminal point of g. So I would go straight up like this, so that's my f plus g.

Now to describe the resultant means, to find its magnitude, find the length of this vector. And I think it's pretty clear that the direction is going to be vertical, but anyway we can talk about that in a second.

How are we going to find the magnitude of this force? Well, let's observe that we can find this angle if we extend. Let's extend this vector just a little bit in the backwards direction. Notice these two angles of 75 degrees add up to 150 and together this is going to make a straight angle. So you're going to need another 30 degrees to make that straight angle.

Let's observe that these two vectors, are exactly the same vector. These are parallel lines, they intersect a transversal. That means this is going to be a 30 degree angle.

So I have a triangle that has two sides of length 500, an angle of 30 degrees, and I need to find this length, so I can do that using the law of cosines. So the length of f plus g² is 500² plus 500² minus 2 times the product of 500 and 500 times the cosine of 30 degrees.

So I'm going to use my calculator and get this, and it's actually going to be an approximation. Let's see 500² plus 500² minus 2 times 500 times 500 times cosine 30. I get 666,987, so my approximation for f plus g is the square root of that, 258.8 Newtons.

Well, if we assume that these two angles were being made with a vertical, then this direction is vertical, and so we can say the direction of the resultant is straight up. It makes sense. They would have to be straight because you're supporting a light by working against gravity which acts straight down. So the magnitude of the force is 258.8 Newtons, and it acts vertically.

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###### 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!”

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