Finding magnitude and direction of equilibrant.


by Obama
Tags: direction, equilibrant, magnitude
Obama
Obama is offline
#1
Feb8-09, 02:17 PM
P: 13
1. The problem statement, all variables and given/known data
Find the magnitude and direction of the equilibrant of each of the following foces:

a) forces of 32N and 48N acting at an angle of 90 degress to each other
b) forces of 16N and 10N acting at an angle of 19 degrees to each other

2. Relevant equations
Sine Law, Cosine law, Soh Cah Toa, Pythagorean.


3. The attempt at a solution

a) Because there is a 90 degree angle, to find the missing side created to make a triangle with the two forces, I would use the pythagorean theorum. The square root of 32^2+48^2 is 57.7 Newtons. This I know is correct. However, when trying to find the direction, I've become puzzled. Since it is a right angled triangle, I used soh cah toa, in which: tanx=(32/48), x=34 degrees. So the direction would be 34 degrees to 32N, correct? The answer page is 146 to 48 N. I'm not sure if there can be two correct answers, as I have found that 180 subtract my original answer of 34 would give me 146, the "correct" answer. Could someone explain this to me?

b) Since there is no right angle, I am using the Cosine law. a^2=10^2 + 16^2-2(10)(16)cos170 (By creating a parallelogram, I found the angle opposite to a is 180-10 degrees). The answer I get is 25.9N. Again, this is correct, and again, I've no clue why my direction is not the same as the answer page.

Using the sine law, sinx/16N = sin170/26. The answer is 6.1 degrees to 16N. The back of the book says 174 degrees to 10N. Again, I have found that 180 subtract my answer of 6.1 degrees would give me 174, the correct answer. I have absolutely no idea why I must subtract by 180 to find the answer. Please, anyone care to explain to me?

This is part of the Calculus and Vectors course, but since this is vectors, I figured it would be more appropriate to post it in the Physics section,thanks.

EDIT: Actually, my teacher advised the physics method is different, so if somone can move this to the Calculus section, you are appreciated.
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LowlyPion
LowlyPion is offline
#2
Feb8-09, 09:14 PM
HW Helper
P: 5,346
There are several ways to add them.

One is to resolve them into components i,j and add the components and resolve the Resulting vector.

The other head to tail addition, which makes the parallelogram like you were doing. It all should come out to the same result.
I am using the Cosine law. a^2=10^2 + 16^2-2(10)(16)cos170
On this one I don't understand your angle.
If they are 19 from each other I would think the angle was 161 if you were adding 1 tail to head of the other.


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