A force of 369 lbs makes an angle of 18 degrees 20 minutes with a force 427 lbs. Find the angle made by the equilibrant with the 369 lbs force.
I know I can find this using the law of cosines, but I think my teacher wants me to use some vector thing, but I do not know how to relate the two vectors to the equilibrant. Is it like u-v?
2 answers
Hmmm. use the 427 Lbs at 0 degrees, then the cos18'20" * 369 is the force in the direction of zero degrees, and 369*sin18'20"is the force 90 degrees to that. Now add these two to see the resulatant, and then add180 degrees to get the equilibrant.
Well, either way it is a "vector thing.
The "equilibrant" is equal and opposite to the "resultant" or sum of the two vectors. In other words it is the vector needed to cancel them.
I will find the resultant, then we can do minus signs.
I am going to say we have 369 pounds along the x axis and 427 pounds at 18 1/3 degrees above the x axis.
In vector notation that is
Rx = 369 + 427 cos 18.33
Ry = 427 sin 18.33
or
Rx = 369 + 405 = 774
Ry = 134
Our equilibrant has components then
Ex = -774
Ey = -134
That is in the third quadrant at
tan A = 134/774 where A is below the -x axis
A = 9.82 degrees below -x axis
or
189.8 degrees counterclockwise from x axis where our original 369 lb force was
The "equilibrant" is equal and opposite to the "resultant" or sum of the two vectors. In other words it is the vector needed to cancel them.
I will find the resultant, then we can do minus signs.
I am going to say we have 369 pounds along the x axis and 427 pounds at 18 1/3 degrees above the x axis.
In vector notation that is
Rx = 369 + 427 cos 18.33
Ry = 427 sin 18.33
or
Rx = 369 + 405 = 774
Ry = 134
Our equilibrant has components then
Ex = -774
Ey = -134
That is in the third quadrant at
tan A = 134/774 where A is below the -x axis
A = 9.82 degrees below -x axis
or
189.8 degrees counterclockwise from x axis where our original 369 lb force was
0 answers