25:[["$","title","0",{"children":"Find the perpendicular components of a force"}],["$","meta","1",{"name":"description","content":"Class 9 FBISE Physics - Turning effect of forces - Find the perpendicular components of a force"}],["$","meta","2",{"name":"author","content":"Editor"}],["$","link","3",{"rel":"manifest","href":"https://classnotes.xyz/site.webmanifest","crossOrigin":"$undefined"}],["$","meta","4",{"name":"keywords","content":"Physics, Turning effect of forces"}],["$","meta","5",{"name":"creator","content":"classnotes"}],["$","meta","6",{"name":"robots","content":"index,follow"}],["$","meta","7",{"name":"application/ld+json","content":"{\"@context\":\"https://schema.org\",\"@type\":\"QAPage\",\"mainEntity\":{\"@type\":\"Question\",\"name\":\"

Find the perpendicular components of a force of 50 N making an angle of 30⁰ with an x-axis. (43.3 N, 25 N)

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Solution:

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Force F = 50 N

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Angle θ = 30⁰

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$F_{x}$ = ? and $F_{y}$ = ?

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$F_{x}$ = F cos θ

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$F_{x}$= $50 \\\\timescos 30$

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= $50 N\\\\times 0.866$ (˙.˙ cos 30⁰ = 0.866)

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$F_{x}$ = 43.3 N

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Similarly, $F_{y}$ = F sin θ

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$F_{y}$= $50 \\\\times 0.5$(˙.˙ sin 30⁰ = 0.5)

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$F_{y}$= 25 N

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