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The Resultant of Two Forces - Concept
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When vectors represent forces, their sum is called the resultant. The resultant of two forces can be found using the methods for adding vectors when the vectors are a geometric representation. When using methods for the algebraic representation to find **the resultant of two forces**, it can be helpful to understand the components of a force.

I want to talk about a special application of adding vectors and that's the resultant of two forces. When two forces act on the same point or object their sum is often called their resultant, the resultant of the two forces, so I want to determine the resultant or sum of two forces f and g, so imagine these two forces f and g are acting on some point or object and the magnitude of f is 500 newtons, that's the unit of force. Magnitude of g is 125 newtons. Both forces are acting on the same direction, what's the resultant? Oh! Remember the resultant is also going to be a vector so you should describe it in terms of its magnitude and its direction.

Now here specifics about the direction aren't really given so we can just say is the resultant is in the same direction as f and g or the opposite direction or whatever something like that. Anyway, when we add vectors remember we like to get them head to tail or we use the parallelogram rule. Let's get them head to tail, what I'll do is I'll take g and translate it over so that it's at the head of force f, okay this way, and here's g. Now when I draw the resultant, the resultant will start at the tail of f and end at the head of g and this will be, this vector will be the resultant f+g, so this is f+g and you can kind of tell that f+g is acting in the same direction as f and g. And what's the magnitude? The magnitude of f+g, since they're both pointing in the same direction, it's just this total length it's 500+125, 625 newtons.

Now what about this situation in part b? f and g are acting on opposite directions. We still add vectors the same way. I'm going to take this vector and translate it so that its tail is at the head of vector f, so if we draw that, so here is vector g and notice the resultant is going to start at the at the tail of f and will end at the head of g so this is the resultant f+g. You will notice that f+g is also in the same direction as f, same direction as f, but it's the opposite direction of g. And the magnitude of f+g, because the two vectors are on opposite directions, you're going to have 500-125, 500 newtons minus 125 newtons that's 375 newtons.

Okay those, those were too easy cases. What if vectors f and g are perpendicular to one another? So in this instance you can you could use the head to tail method but you'll often see in a Physics class the teacher will use the parallelogram method. In this case, the parallelogram is a rectangle. Now what you're basically doing here is you're drawing a vector that's parallel that's basically equal to f and another one that's equal to g and you create a parallelogram. The resultant starts at the common endpoint the initial point of the two vectors and it goes to the opposite corner of the parallelogram, so this is going to be your resultant, and I'll draw in the head so it's clear so this is your f+g. Now what's the direction? What's the magnitude? Well the magnitude is easiest because the length of this leg is 500, the length of this is 125 newtons, so the magnitude squared equals 500 squared or forget the units for now but in the end the units of the magnitude are going to be newtons, 500 squared plus 125 squared, so let's get out our calculators, 500 squared plus 125 squared and then you check the square root and you get 515.4 newtons, this is an approximation 515.4 newtons, so that's the magnitude.

Now what about the direction? Why don't we find this angle and we can say that in the end that the direction of f+g is so many degrees above the horizontal. Let's find what theta it is. Now you can see that in this traingle this is a right triangle the tangent of theta is 125 over 500 the newtons will cancel and you get one quarter, so theta is inverse tangent of one quarter. Again we get our calculator now I want to make sure I'm in degree mode because I want an answer in degrees. Okay, inverse tangent of one quarter 14 degrees approximately 14 degrees so this means that the direction of the force is 14 degrees above the horizontal, the magnitude is 515.4 newtons.

Remember when you're describing a force or any vector give magnitude and direction.

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