source : yahoo.com
Find the magnitude of the resultant force and the angle it makes with the positive x-axis.?
first break down the vectors into x and y components using cos and sin.
a = (200cos(60), 200sin(60) b = (300cos(180), 300sin(180)) y component of b will be zero
next add these two vectors
a+b = < 200cos(60) + 300cos(180), 200sin(60) + 0 > = <-200, 173.2>
the force is the magnitude of this vector
Force = sqrt[ -200^2 + 173.2^2 ] = 264.6N
To find the angle it helps to draw a+b and see where it is. We can see the vector points in the upper left quadrant (i don’t know the numbers) so we can now use the x and y components along with the magnitude to draw a right triangle with the hypotenuse being the magnitude. The angle between the Y axis and the hypotenuse can be calculated using sin.
sin(x) =[x-component/hypotenuse]=200/264.6= 49.1
since this angle is between the Y-axis and the vector, and the question asks for the angle between the X-axis and the vector we simply add 90 degrees.
49.1 + 90 = 139.1
sorry i could not draw the picture. it really helps to understand the angle part if you see the triangle.
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