Question

For the situation shown in the figure, find the values of Ft1 and Ft2 if the object's weight is 600 newtons.​

Answer 1

Given:

Value of weight of object is 600 N.

To find:

The value of F_(T1) and F_(T2) ?

Calculation:

In these kind of questions, we need to apply Lami's Theorem to solve:

According to the Lami's Theorem:

$\therefore \dfrac{600}{ \sin( {180}^{ \circ} - {50}^{ \circ} ) } = \dfrac{F_{T2}}{ \sin( {90}^{ \circ} ) } = \dfrac{F_{T1}}{ \sin( {90}^{ \circ} + {50}^{ \circ} ) }$

$\implies \dfrac{600}{ \sin( {50}^{ \circ} ) } = \dfrac{F_{T2}}{ \sin( {90}^{ \circ} ) } = \dfrac{F_{T1}}{ \cos( {50}^{ \circ} ) }$

$\implies \dfrac{600}{ \sin( {50}^{ \circ} ) } = \dfrac{F_{T2}}{1} = \dfrac{F_{T1}}{ \cos( {50}^{ \circ} ) }$

So, we can say:

$\therefore \: \dfrac{600}{ \sin( {50}^{ \circ} ) } = \dfrac{F_{T2}}{1}$

$\implies \: F_{T2} = \dfrac{600}{ \sin( {50}^{ \circ} ) }$

$\implies \: F_{T2} = \dfrac{600}{0.766}$

$\boxed{ \implies \: F_{T2} = 783.28 \: N }$

We can also say:

$\therefore \: \dfrac{600}{ \sin( {50}^{ \circ} ) } = \dfrac{F_{T1}}{ \cos( {50}^{ \circ} ) }$

$\implies \: F_{T1} = \dfrac{600}{ \{ \frac{ \sin( {50}^{ \circ} ) }{ \cos( {50}^{ \circ} ) } \} }$

$\implies \: F_{T1} = \dfrac{600}{ \tan( {50}^{ \circ} )}$

$\implies \: F_{T1} = \dfrac{600}{1.19}$

$\boxed{\implies \: F_{T1} = 504.2 \: N }$

$\star$ Hope It Helps.

Answer 2

Applying Lami's Theorem,

$\sf \dfrac{600}{sin(180^o - 50^o)} = \dfrac{F_{T2}}{sin90^o} = \dfrac{F_{T1}}{sin(90^o + 50^o)}$

Using the identity,

$\sf \quad \quad \bullet sin(\pi - \theta) = sin\theta$

$\sf \quad \quad \bullet sin\bigg(\dfrac{\pi}{2} + \theta\bigg) = cos\theta$

$\sf \longrightarrow \dfrac{600}{sin50^o} = \dfrac{F_{T2}}{1} = \dfrac{F_{T1}}{cos50^o}$

So,

$\longrightarrow \sf \dfrac{600}{sin50^o} = F_{T2}$

$\sf \longrightarrow F_{T2} = \dfrac{600}{0.766}$

$\longrightarrow \underline{\underline{\sf{\green{ F_{T2} \approx 783.289\:N}}}}$

Now,

$\sf \dfrac{F_{T2}}{1} = \dfrac{F_{T1}}{cos50^o}$

$\longrightarrow \sf F_{T1} = 783.289 \times cos50^o$

$\longrightarrow \sf F_{T1} = 783.289 \times 0.643$

$\longrightarrow \underline{\underline{\sf{\green{F_{T1} \approx 503.488\:N}}}}$