Question

What is the resultant force of the free body diagram?

Answer 1

1) Magnitude of the net force: 20.8 N

2) Angle: $35.2^{\circ}$ clockwise from positive x-direction

Explanation:

1)

First of all, we need to find the components of the resultant force along the horizontal and vertical direction.

We observe that there are two forces acting along the horizontal direction:

$F_1=33 N\\F_2 = 16 N$

We notice that they act in opposite directions, so the net force in the horizontal direction is the difference between these two forces:

$F_x =F_1-F_2=33-16=17 N$ (to the right)

Instead, there is only one force acting in the vertical direction, so the net force in the vertical direction is:

$F_y=F_3=12 N$ (downward)

Therefore, the magnitude of the net force can be found by using Pythagorean's theorem:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{17^2+12^2}=20.8 N$

2)

Now we have to find the angle of the resultant net force. We can do it by using the following equation:

$tan \theta = \frac{F_y}{F_x}$

where

$\theta$ is the angle, measured as clockwise from the positive x-direction (because the vertical net force is downward)

$F_y = 12 N$ is the vertical net force

$F_x=17 N$ is the horizontal net force

Solving, we find:

$tan \theta=\frac{12}{17}=0.706$

And so the angle is

$\theta=tan^{-1}(0.706)=35.2^{\circ}$

In a direction down with respect to the right.

Learn more about forces:

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About addition of vectors:

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