Question

A telephone pole has three cables pulling as shown from above, with F⃗ 1=(300.0iˆ+500.0jˆ) , F⃗ 2=−200.0iˆ , and F⃗ 3=−800.0jˆ . A) Find the net force on the telephone pole in component form. B) Find the magnitude an direction of this net force.

Answer 1

A) Net force in component form: $F=100.0i-300.0j$

B) Magnitude of the net force: 316.2, direction: $-71.6^{\circ}$

Explanation:

A)

The three forces given in this problem are:

$F_1=300i+500j$

$F_2=-200i$

$F_3=-800 j$

The three forces are given in component form, where the components with unit vector i is the component along the x-direction, while the components with unit vector j is the component along the y-direction.

In order to find the net force in component form, we just need to add the components of the three forces along each direction. Therefore:

- Along the x-direction:

$F_x = F_{1x}+F_{2x}+F_{3x}=300+(-200)+0=100$

- Along the y-direction:

$F_y=F_{1y}+F_{2y}+F_{3y}=500+0+(-800)=-300$

So, the net force in component form is

$F=100.0i-300.0j$

B)

The magnitude of a vector F is given by Pythagorean's theorem:

$|F|=\sqrt{F_x^2+F_y^2}$

where in this problem,

$F_x=100$ is the x-component

$F_y=-300$ is the y-component

Substituting,

$|F|=\sqrt{(100)^2+(-300)^2}=316.2$

The direction instead is given by

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{-300}{100})=-71.6^{\circ}$

where the negative sign means the direction is below the positive x-axis.

Learn more about vector addition:

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