A slob is subjected to two force -F1 and -F2 of same magnitude F as shown in the figure -Force -F2 is in XY- plane while force -F1 acts along z-axis at the point 2-i-3-j- The moment of these forces about point O will be -

The correct option is D (3î 2ĵ+3k̂)F →F_1=F2( î)+f√(3)2( ĵ) →r_1=0î+6ĵ →τ_F_1=→r_1×→F_1=3Fk̂ Torque for F_2 force #160; →F_2=Fk ...

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Byju's Answer

Standard XII

Physics

Intro to Torque

A slob is sub...

Question

A slob is subjected to two force $_{1}$ and $_{2}$ of same magnitude $F$ as shown in the figure .Force $_{2}$ is in $X Y -$ plane while force $_{1}$ acts along $z -$ axis at the point $(2 \to i + 3 \to j)$ . The moment of these forces about point O will be :

(3^i - 2^j - 3^k) F

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(3^i + 2^j + 3^k) F

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(3^i + 2^j - 3^k) F

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(3^i - 2^j + 3^k) F

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Solution

The correct option is D $(3^i - 2^j + 3^k) F$
$_{1} = \frac{F}{2} (-^i) + \frac{f \sqrt{3}}{2} (-^j)$

$_{1} = 0^i + 6^j$

$_{F_{1}} =_{1} \times_{1} = 3 F^k$

Torque for $F_{2}$ force

$_{2} = F^k$

$_{2} = 2^i + 3^j$

$_{F_{2}} =_{2} \times_{2} = 3 F^i + 2 f (-^j)$

$_{n e t} =_{F_{1}} +_{F_{2}}$

$= 3 F^i + 2 F (-^j) + 3 F (^k)$

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A slob is subjected to two force →F1 and →F2 of same magnitude F as shown in the figure .Force →F2 is in XY− plane while force →F1 acts along z−axis at the point (2→i+3→j). The moment of these forces about point O will be :

The correct option is D (3^i−2^j+3^k)F→F1=F2(−^i)+f√32(−^j)→r1=0^i+6^j→τF1=→r1×→F1=3F^kTorque for F2 force →F2=F^k→r2=2^i+3^j→τF2=→r2×→F2=3F^i+2f(−^j)→τnet=→τF1+→τF2=3F^i+2F(−^j)+3F(^k)

A slob is subjected to two force $_{1}$ and $_{2}$ of same magnitude $F$ as shown in the figure .Force $_{2}$ is in $X Y -$ plane while force $_{1}$ acts along $z -$ axis at the point $(2 \to i + 3 \to j)$ . The moment of these forces about point O will be :