Rotational Motion

JEE Main level, single correct, medium

125 of 280

Question 125 of 280

A slab is subjected to two forces F1→\overrightarrow{\mathrm{F}_{1}} and F2→\overrightarrow{\mathrm{F}_{2}} of same

magnitude FF as shown in the figure. Force F2→\overrightarrow{\mathrm{F}_{2}} is in XY-

plane while force F1\mathrm{F}_{1} acts along zz -axis at the point

(2i⃗+3j⃗)(2 \vec{i}+3 \vec{j}). The moment of these forces about point O\mathrm{O} will be:

Show answer and solution

Answer: (A) (3i^−2j^+3k^)(3 \hat{i}-2 \hat{j}+3 \hat{k}) F

Given, F→1=F2(−i^)+F32(−j^)\overrightarrow{\mathrm{F}}_{1}=\frac{\mathrm{F}}{2}(-\hat{\mathrm{i}})+\frac{\mathrm{F} \sqrt{3}}{2}(-\hat{\mathrm{j}})

r→1=0i^+6j^\overrightarrow{\mathrm{r}}_{1}=0 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}

Torque due to F1\mathrm{F}_{1} force

τ⃗F1=r→1×F→1=6j^×(F2(−i^)+F32(−j^))=3 F(k^)\vec{\tau}_{\mathrm{F}_{1}}=\overrightarrow{\mathrm{r}}_{1} \times \overrightarrow{\mathrm{F}}_{1}=6 \hat{\mathrm{j}} \times\left(\frac{\mathrm{F}}{2}(-\hat{\mathrm{i}})+\frac{\mathrm{F} \sqrt{3}}{2}(-\hat{\mathrm{j}})\right)=3 \mathrm{~F}(\hat{\mathrm{k}})

Torque due to F2\mathrm{F}_{2} force

τ⃗F2=(2i^+3j^)×Fk^=3 Fi^+2 F(−j^)τ⃗net=τ⃗F1+τ⃗F2=3Fi+2 F(−j^)+3 F(k^)=(3i^−2j^+3k^)F\begin{aligned} \vec{\tau}_{\mathrm{F}_{2}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}) \times & \mathrm{F} \hat{\mathrm{k}}=3 \mathrm{~F} \hat{\mathrm{i}}+2 \mathrm{~F}(-\hat{\mathrm{j}}) \\ \vec{\tau}_{\mathrm{net}}=\vec{\tau}_{\mathrm{F}_{1}}+\vec{\tau}_{\mathrm{F}_{2}} &=3 \mathrm{Fi}+2 \mathrm{~F}(-\hat{\mathrm{j}})+3 \mathrm{~F}(\hat{\mathrm{k}}) \\ &=(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \mathrm{F} \end{aligned}

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