Question #626cb

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Question #626cb

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Answer 1 · 2018-01-23T11:50:25

$the resultant force is 19.08 Newtons and its direction makes 33$ $"the resultant force is 19.08 Newtons and its direction makes 33"$ $degree angle with positive x-axis.$ $"degree angle with positive x-axis."$

Explanation:

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$resultant force= - - \to O C$ $"resultant force="vec (OC)$
$- - \to O C = - - \to O D + - - \to D C$ $vec(OC)=vec(OD)+vec(DC)$

$resultant force= \to F$ $"resultant force="vec F$
$\to F = \to u + \to l + \to k$ $vec F=vec u+ vec l+vec k$
$u = 10 N$ $u=10N$
$l = 12 \cdot cos (60) = 12 \cdot \frac{1}{2} = 6 N$ $l=12*cos(60)=12*1/2=6 N$

$- - \to O D = u + l = 10 + 6 = 16 N$ $vec(OD)=u+l=10+6=16N$

$k = 12 \cdot sin 60 = 12 \cdot \frac{\sqrt{3}}{2} = 6 \sqrt{3} N$ $k=12*sin 60=12*sqrt 3/2=6 sqrt 3N$
$- - \to D C = k = 6 \sqrt{3} N$ $vec(DC)=k=6 sqrt 3 N$
$F = \sqrt{{(O D)}^{2} + {(D C)}^{2}}$ $F=sqrt((OD)^2+(DC)^2)$

$F = \sqrt{16^{2} + {(6 \sqrt{3})}^{2}}$ $F=sqrt(16^2+(6sqrt 3)^2)$

$F = \sqrt{256 + 108}$ $F=sqrt(256+108)$

$F = \sqrt{364}$ $F=sqrt(364)$

$F = 19.08 N$ $F=19.08 N$

$tan β = \frac{D C}{O D}$ $tan beta=(DC)/(OD)$

$tan β = \frac{6 \sqrt{3}}{16} = 0.64951905$ $tan beta=(6sqrt 3)/16=0.64951905$

$β = 33 d e g$ $beta=33 deg$

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Question #626cb

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