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Inertia

Forces F1 and...

Question

Forces $F_{1} and F_{2}$ act on a point mass in two mutually perpendicular directions. The resultant force on the point mass will be

F_{1} - F_{2}

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F_{1} + F_{2}

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\sqrt{F_{1}^{2} + F_{2}^{2}}

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F_{1}^{2} + F_{2}^{2}

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Solution

The correct option is C $\sqrt{F_{1}^{2} + F_{2}^{2}}$
$A = F_{1}, B = F_{2}$
$Θ = 90 °$
using,
$∣ ∣ \to R ∣ ∣ = \sqrt{A^{2} + B^{2} + 2 A B cos Θ}$
put the values, we get
$F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 90 °} = \sqrt{F_{1}^{2} + F_{2}^{2}}$

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Inertia

Standard XII Physics

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Forces F1 and F2 act on a point mass in two mutually perpendicular directions. The resultant force on the point mass will be

The correct option is C √F21+F22A=F1, B=F2 Θ=90° using, ∣∣→R∣∣=√A2+B2+2ABcosΘ put the values, we get F=√F21+F22+2F1F2cos90°=√F21+F22

Forces $F_{1} and F_{2}$ act on a point mass in two mutually perpendicular directions. The resultant force on the point mass will be

The correct option is C $\sqrt{F_{1}^{2} + F_{2}^{2}}$
$A = F_{1}, B = F_{2}$
$Θ = 90 °$
using,
$∣ ∣ \to R ∣ ∣ = \sqrt{A^{2} + B^{2} + 2 A B cos Θ}$
put the values, we get
$F = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 90 °} = \sqrt{F_{1}^{2} + F_{2}^{2}}$