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Magnitude and Direction Force Problem 
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#1
Nov2209, 11:52 PM

P: 13

1. The problem statement, all variables and given/known data
Use the cosine and sine rules to determine the magnitude and direction of the resultant of a force of 11 kN acting at an angle of 50 degrees to the horizontal and a force of 8 kN acting at an angle of 30 degrees to the horizontal. 2. Relevant equations Law of Sines: (a/sinA) = (b/sinB) = (c/sinC) Law of Cosines: b^2 = a^2 + c^2  2*a*c*cosB 3. The attempt at a solution Ok, so I actually already solved for the magnitude to be 14.29783742 kN. But the other blank for the question is for "degrees to the horizontal". I may not be understanding the term correctly but I have no real idea of how to solve for this. 


#2
Nov2309, 06:08 AM

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Hi Dr Meow! Welcome to PF!
(try using the X^{2} tag just above the Reply box ) Find the angle between the resultant and the 11kN force … then subtract that angle from 50º (which is the angle between the 11kN force and the horizontal) to get the angle between the resultant and the horizontal. 


#3
Nov2309, 08:09 PM

P: 13

Isn't the resultant just the horizontal line? I'm sorry I did this problem a while ago and I don't remember how I did it. And the reason for the accuracy is because I'm doing this problem on WebAssign and you can't round to put in the right answer.



#4
Nov2409, 02:22 AM

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Magnitude and Direction Force Problem
(and I don't make it 14.29783742) 


#5
Nov2409, 10:12 PM

P: 13

I'm sorry, I didn't understand what you really meant by the resultant in this case. When you said solve for the resultant.



#6
Nov2509, 02:20 AM

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Hi Dr Meow!
(just got up …) I thought you knew that, because you found its magnitude. Had you forgotten what you did? btw, I didn't say "solve for the resultant", I said … 


#7
Jan1010, 08:55 PM

P: 13

Hey sorry it's taken me so long to reply I had forgotten about this. To be honest I did forget how I found the magnitude. And I'm still not sure because it's been even longer. Is there a way you could help me do just the degrees to the horizontal?
Thanks 


#8
Jan1110, 03:26 AM

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Hi Dr Meow! Happy new year!
Show us what you've done, and where you're stuck, and then we'll know how to help. 


#9
Jan1110, 09:56 PM

P: 13

Ok. So here's what I've done: To get the magnitude I drew a diagram of the 30 degree angle downwards and the 50 degree angle upwards. Then combined them to make a total of 80 degrees and used law of cosines to find the hypotenuse of the total triangle created and found that to be 14.29, which is correct. What I'm unsure of is exactly how to find the "degrees to the horizontal".



#10
Jan1110, 10:21 PM

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If α is the angle of the resultant of P and Q with P ( here it is 8 kN), then with a simple geometry you can show that
tan(α) = Q*sinθ/(P + Q*cosθ), where θ is the angle between P and Q. Then (α  30) will be the angle of the resultant with horizontal. 


#11
Jan1110, 11:51 PM

P: 13

So would it end up being tan(α) = (11)*sin(80)/[(8 + 11)*cos(80)]? Then (α  30) Because when I computed that all out it ended up to be the wrong answer.



#12
Jan1210, 04:31 AM

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There is another method. Find the vertical and horizontal components of 11 kN and 8 kN. If φ is the angle between the resultant and horizontal, then tanφ = ΣFy/ΣFx 


#13
Jan1210, 07:54 AM

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This is wrong. It should be tan(α) = (11)*sin(80)/[(8 + 11*cos(80)]? 


#14
Jan1210, 11:30 PM

P: 13




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