Question

please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

Answer 1

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

     = 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

    = 9.58 m

Similarly, Horizontal component of vector C,

$C_x$  = C[Cos(60)]

     = 6[Cos(60)]

     = $\frac{6}{2}$

     = 3 m

$C_y=6[\text{Sin}(60)]$

    = 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

   = $\sqrt{(4.22)^2+(4.38)^2}$

   = $\sqrt{17.81+19.18}$

   = 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

                  = $\frac{R_y}{R_x}$

                  = $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

             = 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.