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Standard IX

Physics

Second Law of Motion

Two forces ...

Question

Two forces $F_{1} = 1 N$ and $F_{2} = 2 N$ act along the lines $x = 0$ and $y = 0$ , respectively. Then find the resultant of forces.

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Solution

$F_{1} = 1 N F_{2} = 2 N$
angle between both the forces is $90^{0}$ .
$R = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 90^{0}}$
$R = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$
$\to F = 2^i +^j$
$t a n θ = \frac{1}{2} \Rightarrow θ = {tan}^{- 1} (\frac{1}{2})$ direction of force resultant.

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Two forces F1=1N and F2=2N act along the lines x=0 and y=0, respectively. Then find the resultant of forces.

F1=1NF2=2Nangle between both the forces is 900.R=√F21+F22+2F1F2cos900R=√12+22=√5→F=2^i+^jtanθ=12⇒θ=tan−1(12) direction of force resultant.

Two forces $F_{1} = 1 N$ and $F_{2} = 2 N$ act along the lines $x = 0$ and $y = 0$ , respectively. Then find the resultant of forces.

$F_{1} = 1 N F_{2} = 2 N$
angle between both the forces is $90^{0}$ .
$R = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 90^{0}}$
$R = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$
$\to F = 2^i +^j$
$t a n θ = \frac{1}{2} \Rightarrow θ = {tan}^{- 1} (\frac{1}{2})$ direction of force resultant.