Question

Don't understand questions 1-4!

Answer 1

1ª

a)
$|FR| = \sqrt{ Fr_{x}^2 + Fr_{y}^2 } = \sqrt{ 2^2 + 3.1^2 } = \sqrt{4+9.61} = \sqrt{13,61}$
$|FR| = 3.689173349$
$\boxed{\boxed{|FR| \approx 3.7N}}$

magnitude: 3.7 N
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$\Theta = tg^{-1} | \frac{ Fr_{x} }{ Fr_{y} } |$
$\Theta = tg^{-1} ( \frac{2}{3.1} )$
$\Theta = 32.82854179$
$\boxed{\boxed{\Theta \approx 33^0anticlockwise/horizontal}}$

b)

$|FR| = \sqrt{ Fr_{x}^2 + Fr_{y}^2 } = \sqrt{ 16^2 + 6^2 } = \sqrt{256+36} = \sqrt{292}$
$|FR| = 17.08800749$
$\boxed{\boxed{|FR| \approx 17.1N}}$

magnitude: 17.1 N
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$\Theta = tg^{-1} | \frac{ Fr_{x} }{ Fr_{y} } |$
$\Theta = tg^{-1} ( \frac{6}{16} )$
$\Theta = 20.55604522$
$\boxed{\boxed{\Theta \approx 20.6^0anticlockwise/horizontal}}$ or $\boxed{\boxed{\Theta \approx 21^0anticlockwise/horizontal}}$

c) 

$|FR| = \sqrt{ Fr_{x}^2 + Fr_{y}^2 } = \sqrt{ (3-2)^2 + 1^2 } = \sqrt{1^2+1^2} = \sqrt{2}$
$|FR| = 1.414213562$
$\boxed{\boxed{|FR| \approx 1.4N}}$

magnitude: 1.4 N
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Watch the picture:
We have: 45º (isosceles) below horizontal (to 3N and 1N).
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4ª

a)

$Cos40^0 = \frac{y}{350}$

$0.7660444431 = \frac{y}{350}$

$y = 0.7660444431*350$

$y= 268.1155551$

$\boxed{\boxed{y \approx 268 N}}$

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b)

In opposite directions

$|FR| = \sqrt{ Fr_{x}^2 - Fr_{y}^2 } = \sqrt{ 350^2 - 268^2 } = \sqrt{122500 - 71824} = \sqrt{50676}$
$|FR| = 225.1133048$
$\boxed{\boxed{|FR| \approx 225N}}$