Question

Juan is helping his mother rearrange the living room furniture. Juan pushes on the armchair with a force of 30.0 N directed at an angle of 25.0° above a horizontal line while his mother pushes with a force of 60.0 N directed at an angle of 35.0° below the same horizontal.What is the magnitude of the vector sum of these two forces?What is the direction of the vector sum of these two forces? If the direction is below the horizontal, enter a negative value and if the direction is above the horizontal, enter a positive value.

Answer 1

Given:

For Juan:

Force, F = 30.0 N

θ = 25.0° above a horizontal line

For his mother:

Force, F = 60.0 N

θ = 35.0° below a horizontal line.

Let's solve for the following:

• (a) What is the magnitude of the vector sum of these two forces?

For Juan, the force(x-component) will be:

Fx = 30 cos(25)i + 30 sin(25)j

For his mother, the force(y-component) will be:

Fy = 60 cos(35)i - 60 sin(35)j

Now, let's find the net force:

Fnet = Fx + Fy

Fnet = 30 cos(25)i + 30 sin(25)j + 60 cos(35)i - 60 sin(35)j

Fnet = 30 cos(25)i + 60 cos(35)i + 30 sin(25)j - 60 sin(35)j

Fnet = 76.338i - 21.736j

To find the magnitude of the vector sum of these two forces, we have:

$F_{\text{net}}=\sqrt[]{(F_x)^2+(F_y)^2}_{}$

$\begin{gathered} F_{\text{net}}=\sqrt[]{(76.338)^2+(-21.736)^2} \\ \\ F_{\text{net}}=79.37\text{ N} \end{gathered}$

Therefore, the magnitude of the vector sum of these two forces is 79.37 N.

• (b) What is the direction of the vector sum of these two forces?

To find the direction of the vector sum of these two forces, apply the formula:

$\alpha=\tan ^{-1}(\frac{F_y}{F_x})$

Where:

y-component (Fy) = -21.736

x-component (Fx) = 76.338

Thus, we have:

$\begin{gathered} \alpha=\tan ^{-1}(\frac{-21.736}{76.338}) \\ \\ \alpha=-15.89^o \end{gathered}$

Therefore, the direction of the vector sum of these two forces is at -15.89°

ANSWER:

(a) 79.37 N

(b) -15.89°