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I have the question "Resolve the forces acting on the following objects into their vertical and horizontal components and thus find the vertical and horizontal components of their resultant force".

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I know that I have to make a triangle to find the vertical and horizontal forces using trigonometry.

However I am not sure how to form the triangle and the 125 degree angle is confusing me.

Here is my attempt of the triangle:

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So for the vertical I got 2.29 N and for the horizontal I got 3.28 N.

However, the solutions say that the vertical and horizontal should be 0.71 N.

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    $\begingroup$ Draw $x,y$ axes through the origin of the three vectors and then project the oblique vector on the axes. The triangle will (not so) magically appear. $\endgroup$ – Fabio Somenzi Jan 18 '17 at 15:32
  • $\begingroup$ I don't really understand because the vertical and horizontal components are already given. 3.0N and 4.0N. Could you please elaborate. $\endgroup$ – Dan Jan 18 '17 at 15:43
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    $\begingroup$ The way I understand it--that is, the way these problems are typically stated--there are three forces acting on your object. One is horizontal, one is vertical, and the third is neither. You need to decompose the last one into its horizontal and vertical components. These components partially counter the other two forces. See also the answer by @DavidK. $\endgroup$ – Fabio Somenzi Jan 18 '17 at 15:46
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    $\begingroup$ Yes, that's correct. Now you can apply trigonometry to find the two sides from the angle and the hypotenuse. $\endgroup$ – Fabio Somenzi Jan 18 '17 at 16:37
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    $\begingroup$ Those aren’t “vertical and horizontal components... already given.” There are three vectors in the diagram that you have to sum. $\endgroup$ – amd Jan 18 '17 at 19:00
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Sometimes you need to add some lines to the diagram. Since you're looking for horizontal and vertical forces, draw vertical and horizontal lines in both directions from the point where all the force arrows start.

It looks like two of the vectors were meant to be vertical or horizontal already. So when you draw the vertical and horizontal lines, they will be collinear with the arrows depicting those two vectors. A vector perfectly aligned with a horizontal line is its own horizontal component (its vertical component is zero). Likewise a vertical vector is easily split into components (one of which is just zero).

You have one vector where you need trigonometry. It has a $125$-degree angle measured from the vertical line, but you can use that fact to figure out its angle from the horizontal line. That gives you an angle less than $90$ degrees and it's easy to draw a right triangle using that angle.

Alternatively, knowing you have a $125$-degree angle with the vertical line above the vector, and knowing that two angles making a straight line must sum to $180$ degrees, you can figure out the angle the vector makes with the lower part of the vertical line and use that angle to draw a right triangle.

As noted in one of the comments, an important thing to notice is that there really are three vectors in the figure. Two of them happen to be already aligned with the axes, which makes it very easy to identify their components, but that does not make those two vectors somehow be components of the third vector.


When you get more comfortable with components and computing them using the sine and cosine functions (so that you don't need to actually identify triangles to find the components each time), it will be useful to learn how to use the sine and cosine functions to deal directly with angles greater than $90$ degrees. For example, for angles in degrees, $\cos(180 - x) = -\cos(x)$. But that might be a slightly more advanced topic for a later time.

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  • $\begingroup$ Please could you see my edit for the attempt of the triangle. Is this correct ? $\endgroup$ – Dan Jan 18 '17 at 16:07
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    $\begingroup$ @Dan Yes, that looks like a correct triangle for this question. $\endgroup$ – David K Jan 18 '17 at 16:29
  • $\begingroup$ So for the triangle I made I got 2.29 N for the Vertical and 3.38 N for the horizontal. However, the solutions say that the vertical and horizontal should be 0.71 N. How can this be ? $\endgroup$ – Dan Jan 18 '17 at 16:38
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    $\begingroup$ You still have to combine the components you computed with the other forces. The solutions that were given to you are approximate. The horizontal component is slightly off. $\endgroup$ – Fabio Somenzi Jan 18 '17 at 16:42
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    $\begingroup$ I get 3.28 N for the horizontal. But then you have to remember the other two vectors--you're supposed to combine the components of all three vectors to get the result. So for vertical we have 2.29 downward from the 125-degree vector but 3 upward from the vertical vector from the original figure. The 0.71 is the upward component that remains from 3 upward partially canceled by 2.29 down. $\endgroup$ – David K Jan 18 '17 at 16:42

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