Question

I need the find the magnitude of vector A + B .. please see the photo

Answer 1

In order to find the magnitude of the vector sum, proceed as follow:

Calculate the vertical and horizontal components of A, as follow:

$\begin{gathered} A_x=A\cos \theta \\ A_y=A\sin \theta \end{gathered}$

where,

A: magnitude of A vector = 4.00N

θ: direction related to B = 60.0 degrees

Replace the previous values into the expressions for Ax and Ay and simplify:

$\begin{gathered} A_x=4.00N\cdot\cos (60.0)=2.00N \\ A_y=4.00N\cdot\sin (60.0)=3.46N \end{gathered}$

Now, consider that B vector can have only a horizontal component, then, its magnitude is just the value of Bx (it is possible because there is no specifications about the orientation of the vectors in the coordinate system):

$B_x=5.40N$

Next, consider that the sum vector S has the following components:

$\begin{gathered} S_x=A_x+B_x=2.00N+5.40N=7.40N \\ S_y=A_y_{}=3.46N \end{gathered}$

And the magnitude of the sum vector S is:

$S=\sqrt[]{S^2_x+S^2_y}=\sqrt[]{(7.40N)^2+(3.46N)^2}\approx8.17N$

Hence, the magnitude of the vector sum is approximately 8.17N