Resolve two forces at 120º to each other?

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2 Answers
Jan 18, 2016

Resolving the non-horizontal force and adding the components to the horizontal force shows that the answer is 10 N.
(This is one way of solving.)


If we resolve the non-horizontal force into a horizontal and vertical component we can add those to the horizontal force by vector addition to solve the problem.

First resolve the non-horizontal force into horizontal and vertical components:
We can divide the total 120º angel neatly around a vertical line from P. To the right of the vertical is a 90º angle to the horizontal force, therefore from the vertical to the other force is an angle of 30º.
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So the vertical component, #F_V = 10 cos 30 = 8.660 N#.
And the horizontal component, #F_H = 10 sin 30 = 5.0 N#.

Now add the horizontal component to the horizontal force:
The forces are in opposite directions so they subtract. Taking to the right as positive, the resultant of these two forces is #10 – 5.0 = 5.0 N#
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So the total horizontal force is 5.0 N to the right (shown in green in the next diagram).
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The last step is to calculate the magnitude of the resultant of these two forces:
The total horizontal force and the vertical component form a right angled triangle, so we can use Pythagoras’ Theorem.
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#F_R = sqrt(8.660^2 + 5.0^2 ) = 10 N#

Jan 18, 2016

You can use the cosine rule to solve the problem if you rearrange the forces head-to-tail.


If you want a more elegant solution than the one I just provided, well here it is!

We can rearrange the problem so that the two forces are arranged head-to-tail and use the cosine rule to solve for the resultant force.

Rearrange the forces so they are head-to-tail:
The diagram shows the starting arrangement and the new arrangement with the forces head-to-tail. The new arrangement also shows the resultant force (that starts at P and ends where the second 10 N force ends). The 120º angle is shown for reference. We also know that the angle between the 10 N forces is 60º since 180º – 120º = 60º.
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Use the cosine rule to solve for the resultant force:
(Cosine Rule: a² = b² + c² – 2bc cosA)

#(F_R)^2 = 10^2 + 10^2 – 2 × 10 × 10 × cos 60 = 100#
#⇒ F_R = sqrt(100) = 10 N #