Question

the three forces f1 f2 and f3 are acting on particle as shown in figure if particle P is in equilibrium then​

Answer 1

$\maltese\: \underline{\underline{\textsf{AnsWer :}}}\:\maltese$

• F₁ = 4 N
• F₂ = ?
• F₃ = 10 N

By applying Lami's Theorem we have :

$\longrightarrow\:\:\sf \dfrac{a}{\sin \alpha} = \dfrac{b}{\sin \beta} = \dfrac{c}{\sin \gamma}$

$\longrightarrow\:\:\sf \dfrac{F_1}{\sin (\pi - \theta)} = \dfrac{F_2}{\sin (90^{\circ} + 45^{ \circ})} = \dfrac{F_3}{\sin (45^{\circ} + \theta)}$

$\longrightarrow\:\:\sf \dfrac{F_1}{\sin ( {180}^{ \circ} - \theta)} = \dfrac{F_2}{\sin (90^{\circ} + 45^{ \circ})} = \dfrac{F_3}{\sin (45^{\circ} + \theta)}$

$\longrightarrow\:\:\sf \dfrac{4}{\sin ( {180}^{ \circ} - \theta)} = \dfrac{F_2}{\sin (90^{\circ} + 45^{ \circ})} = \dfrac{10}{\sin (45^{\circ} + \theta)}$

$\longrightarrow\:\:\sf \dfrac{4}{\sin \theta} = \dfrac{F_2}{\cos 45^{ \circ}}$

$\longrightarrow\:\:\sf \dfrac{4}{\sin \theta} = \dfrac{F_2}{ \frac{1}{ \sqrt{2} } }$

$\longrightarrow\:\:\sf 4 \times \dfrac{1}{ \sqrt{2} } = F_2 \sin \theta$

$\longrightarrow\:\: \underline{ \boxed{\bold{ F_2 \sin \theta = 2 \sqrt{2} \: N}}}$

Answer 2

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