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Ответ:

Kikilcaro9675

25.07.2020

Physics

The magnitude of the resulting force is 67 lbf.

Explanation:

Taking the east as the positive x direction, and the north as the positive y direction.

The first force points west, this is, in the direction of $-\hat{i}$ , so, is

$\vec{F}_1 = - 20 \ lbf \ \hat{j}$

$\vec{F}_1 = (0 , - 20 \ lbf \)$

For the second force, knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

where $| \vec{A} |$ is the magnitude of the vector and θ the angle with the positive x direction.

So, the second force is

$\vec{F}_2 = 80 \ lbf \ \ ( \ cos(45 \°) \ , \ sin (45 \°) \ )$

$\vec{F}_2 = ( \ 56.5685 \ lbf \ , \ 56.5685 \ lbf \ )$

The net force will be :

$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2$

$\vec{F}_{net} = (0 , - 20 \ lbf \) + ( \ 56.5685 \ lbf \ , \ 56.5685 \ lbf \ )$

$\vec{F}_{net} = ( \ 56.5685 \ lbf \ , \ 56.5685 \ lbf \ - 20 \ lbf )$

$\vec{F}_{net} = ( \ 56.5685 \ lbf \ , \ 36.5685 \ lbf \ )$

To obtain the magnitude, we can use the Pythagorean Theorem

$|\vec{F}_{net}| = \sqrt{F_{net_x}^2 +F_{net_y}^2}$

$|\vec{F}_{net}| = \sqrt{( \ 56.5685 \ lbf \ )^2 +( \ 36.5685 \ lbf \ )^2}$

$|\vec{F}_{net}| = 67.36 \ lbf$

And this is the magnitude we are looking for.

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