Question

In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector C→ has a magnitude of 13.8 m and is angled 20.7° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of B→? State your angle as a positive number.

Answer 1

Answer:

$|B|=27.00425726m$

$\alpha =210.3781372$°

Explanation:

Let's use the component method of vector addition:

$A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844$

Now, we know:

$C_x=A_x+B_x\\\\C_y=A_y+b_y$

So:

$B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774$

Now lets calculate the magnitude of the vector B:

$|B|=\sqrt{(B_x)^{2} +(B_y)^{2} }=27.00425726m$

Finally its angle is given by:

$\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344$°

Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.