Vector problem?

Question

I have a lab write up due tomorrow but I can't figure out the correct angle for the magnitude or vise versa. We had to set up a machine that if you were to get the resultant vector correct, all the weights on the other vectors would reach an equilibrium with the resultant. I'm sure I got the magnitude correct because the weights evened out but the angle the magnitude suggests is incorrect. Here's the problem:

A force of 500g at 60 degrees, a force of 400g at 150 degrees, and a force of 300g at 250 degrees.

I need to find the resultant for that. Any help would be great!

Answer 1

You should sketch these three vectors on paper using an x-y axis coordinate system and then calculate the x-axis and y-axis components of each vector. Don’t try to accurately plot them, just eyeball the lengths and angles to get a rough idea of the magnitudes and directions. Sketch in the projection of each vector on the x-axis and the y-axis. In other words, draw a perpendicular line, parallel to the y-axis, from the tip of the vector the x-axis. The distance from the origin to the intersection of the line with the x-axis is the x-axis component of that vector. It is equal to the magnitude of the vector multiplied by the cosine of the angle the vector makes with the x-axis. The length of the perpendicular line, from the tip of the vector to its intersection with the x-axis, is the y-axis component of that vector. It is equal to the magnitude of the vector multiplied by the sine of the angle the vector makes with the x-axis.

Pay attention to which quadrant a vector is in to obtain the correct signs for the x-axis and y-axis components. In other words, a vector in the first quadrant has both positive x and positive y components, in the second quadrant it has negative x and positive y components, in the third quadrant it has both negative x and negative y components, and in the fourth quadrant it has positive x and negative y components.

Algebraically add the x-axis component of each vector to find the x-axis component of the resultant vector. Remember, the x-axis component is the vector magnitude multiplied by the cosine of the angle the vector makes with the x-axis.

Algebraically add the y-axis component of each vector to find the y-axis component of the resultant vector. Remember, the y-axis component is the vector magnitude multiplied by the sine of the angle the vector makes with the x-axis.

Find the magnitude of the resultant vector by taking the square root of the sum of the squares of the resultant x-axis and y-axis components. Find the direction of the resultant vector by taking the arc tangent of the quotient of the y-axis component divided by the x-axis component. This will be the angle the resultant vector makes with the x-axis.

Use the diagram you drew to determine the signs of the components. Sketch in the resultant vector and verify that it appears to be in the correct quadrant. Sketch in a vector of the same magnitude in the opposite direction (add 180 degrees to the resultant vector) to determine the vector that balances the original three vectors.

500/_60 is in the first quadrant with x-axis component of 500 cos 60 and y-axis component of 500 sin 60.

400/_150 is in the second quadrant with x-axis component of -400 cos 30 and y-axis component of 400 sin 30.

300/_250 is in the third quadrant with x-axis component of –300 cos 70 and y-axis component of –300 sin 70.

The signs and angles are obtained by inspection from the graph of the vectors.

So, for x-axis components we have:

500 cos 60 – 400 cos 30 – 300 cos 70
= 500 (0.500) – 400 (0.866) – 300 (0.342)
= 250 – 346.41 – 102.61
= -199.02

And for the y-axis components we have:

500 sin 60 + 400 sin 30 – 300 sin 70
= 500 (0.866) + 400 (0.500) – 300 (0.940)
= 433.01 + 200 – 281.91
= 351.10

This resultant of the original three vectors is a vector in the second quadrant because x is negative and y is positive. It has magnitude = SQRT (X^2 + Y^2):

SQRT [(-199.02)^2 + (351.10)^2]
= SQRT [39608.9604 + 123271.21
= SQRT 162880.1704
= 403.58

It makes an angle with the x-axis of arc tan (351.10/199.02) = arc tan 1.764
= 60.45 degrees in the second quadrant. The resultant vector is then 403.58 /_ 119.55

To balance this resultant vector and have equilibrium you need a vector in the opposite direction (in the fourth quadrant) of the same magnitude: 403.58 /_299.55

Answer 2

What you need to do is think of each force as a two-component system, one component in the X direction of the Cartesian coordinate system and the other in the Y direction. By summing the X components of the three forces, you will find the average x force. Doing the same with the y forces will get the average Y force. Using the Pythagorean theorem, you can find the average overall force needed to counteract the three forces. To find the direction of the force, you need to find the tangent of the Y and X components. You now have the magnitude and direction of the force. To find the resultant, you need an equal, but opposite force, which would be 180 degrees from the average force.

500g at 60 degrees has an x component of 500g*cosine(60) and a y component of 500g*sine(60) ==> (250, 433)g
400g at 150 degrees ==> (-346, 200)g
300g at 250 degrees ==> (-103, -282)g
The sum of the vectors gives ==> (-199, 351)g
The average force is ==> 403g
The direction is ==> -101 degrees which is also 259 degrees

Therefore, to counteract that weight, you would need a 403g weight at 79 degrees.

I'd check the math on this, but this is how it should be done.

Answer 3

500*cos60+400*cos150+ 300*cos250=-199
500*sin60+400*sin150+ 300*sin250=351.1
sqrt(-199^2+351.1^2)= 403.6g