1 Answer

GreenOwl · Answer 1 · 2018-06-14T19:23:42+0000

1ª

a)

$|FR| = \sqrt{ Fr_(x)^2 + Fr_(y)^2 } = √( 2^2 + 3.1^2 ) = √(4+9.61) = √(13,61)$

|FR| = 3.689173349

$\boxed{\boxed}$

magnitude: 3.7 N
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$\Theta = tg^(-1) | ( Fr_(x) )/( Fr_(y) ) |$

$\Theta = tg^(-1) ( (2)/(3.1) )$

$\Theta = 32.82854179$

$\boxed{\boxed{\Theta \approx 33^0anticlockwise/horizontal}}$

b)

$|FR| = \sqrt{ Fr_(x)^2 + Fr_(y)^2 } = √( 16^2 + 6^2 ) = √(256+36) = √(292)$

|FR| = 17.08800749

$\boxed{\boxedFR}$

magnitude: 17.1 N
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$\Theta = tg^(-1) | ( Fr_(x) )/( Fr_(y) ) |$

$\Theta = tg^(-1) ( (6)/(16) )$

$\Theta = 20.55604522$

$\boxed{\boxed{\Theta \approx 20.6^0anticlockwise/horizontal}}$ or
$\boxed{\boxed{\Theta \approx 21^0anticlockwise/horizontal}}$

c)

$|FR| = \sqrt{ Fr_(x)^2 + Fr_(y)^2 } = √( (3-2)^2 + 1^2 ) = √(1^2+1^2) = √(2)$

|FR| = 1.414213562

$\boxed{\boxed}$

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 = (y)/(350)