Basic Engineering Mechanics (trigonometry)

AI Thread Summary
The discussion revolves around calculating the resultant of two forces applied to a storage bin lid, with given magnitudes of 48N and 60N. The user successfully determined the magnitude of the resultant force as 104N using trigonometric methods and the Pythagorean theorem. However, confusion arose regarding the angle of the resultant relative to the horizontal, with the user initially miscalculating it as 13.3 degrees. Clarifications were made regarding the correct interpretation of angles in relation to the horizontal and the second quadrant, ultimately leading to the correct angle of 86.7 degrees. The conversation highlights the importance of understanding vector angles and their reference points in engineering mechanics.
ellieahlquist
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Homework Statement


Two forces P and Q are applied to the lid of a storage bin as shown (see picture). Knowing that P = 48N and Q = 60N, determine by trigonometry the magnitude and direction of the resultant of the two forces.
IMG_20130903_180133_524_zpse7aada23.jpg



Homework Equations


"The triangle method" where you put each vector tip-to-tail, and then use components of one vector to make a right triangle (see picture).
IMG_20130903_180803_648_zps4cc30f74.jpg

Also, law of sines, law of cosines (??), and sin, cos, tan.
Pythagorean Thm

The Attempt at a Solution



As you can see, I turned the picture somewhat sideways so that Q lays flat.
I used simple trig to find Px = 48cos30o = 41.56 and Py = 48sin30o = 24. From there, I have a right angle and I know that one leg of the triagle is Q+Px and the other is Py. Using Pythagorean theorem, I got 104N, which is the right magnitude for the resultant.

My problem is the angle of the resultant. When I solve for the angle θ (between Q and R) in my picture, I get 13.3.
I used tanθ = 24/101.56
as well as the law of sines (104/sin150) = (48/sinθ)
both yield the 13.3o.
Even adding 13.3 to the 25 degree angle lid and past the 55 degree Q vector, I still don't get the right answer (which is 86.7 degrees, with an angle symbol pointing into the second quadrant, not sure what that means either).

Thanks for any help.
 
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What is the resultant angle with the horizontal? If measured from the second quadrant?
 
voko said:
What is the resultant angle with the horizontal? If measured from the second quadrant?

I guess that's what I'm trying to figure out.
I also find the wording in this problem poor because since the answer is 86.7 pointing to the second quadrant (??), does that mean it's 86.7 from the already 25 degree slant? or from the 1st quadrant x axis? or from where?
 
ellieahlquist said:
I guess that's what I'm trying to figure out.

Are you saying you cannot answer the questions? You have calculated, correctly, the resultant angle with Q. But Q itself is not horizontal. Do you know its angle with the horizontal? What is the resultant angle with the horizontal then?
 
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voko said:
Are you saying you cannot answer the questions? You have calculated, correctly, the resultant angle with Q. But Q itself is not horizontal. Do you know its angle with the horizontal? What is the resultant angle with the horizontal then?

This was my first thought:
The lid is 25 degrees from horizontal plus an additional 55 degrees to get to Q. 25+55=80
So I am at 80 degrees, and if the resultant vector is 13.3 degrees from Q, that's 93.3 degrees.
I'm still missing something because the answer is 86.7 degrees.

But now:
I think I understand what you meant by saying "if measured from the second quadrant." For some reason I didn't take that as from the -x axis. So I have 180 degree horizontal, minus the 25, minus the 55, then I have 100. and 100-13.3 is 86.7 degrees. But I definitely wouldn't have figured that out. Thanks.
 
This is just something to keep in mind. This is done frequently, even if it is confusing. We just prefer angles less than 90 degrees.
 
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