Question

(CENTRE OF MASS, Class 11)

Give proper explanation and answer it correctly.

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Answer 1

Solution:-

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We can see the bar is in equilibrium so

$\rm \: \to \sum F = 0$

$\rm \to \: \Sigma \tau = 0$

Now take 36.9⁰ as 37⁰ and 53.1 as 53⁰ for making easy calculations

=> Horizontal force

$\rm \: T_1 \sin37 ^o = T_2 \sin53 {}^{o}$

Now value of sin37⁰ is 3/5 and sin53⁰ is 4/5

$\rm \: T_1 \times \dfrac{3}{5} = T_2 \times \dfrac{4}{5}$

$\boxed{\rm \: T_1 \dfrac{3}{4} = T_2 }$

=> Vertical force

$\rm \: T _{1} \cos37^{o} + T _{2} \cos53^{o} = w$

$\rm \: T_1 \times \dfrac{4}{5} + T_2 \times \dfrac{3}{5} = w$

now we have

$\to\rm \: T_1 \dfrac{3}{4} = T_2$

So we get

$\rm \: T_1 \times \dfrac{4}{5} + \dfrac{3}{4} T_1 \times \dfrac{3}{5} = w$

$\boxed{\rm \: T_1 \dfrac{5}{4} = w}$

Now we have three unknown value or 2 equation now we using

$\Sigma \tau = 0$

So we know that torques in clockwise is +ve

and torques in anticlockwise is - ve

$\rm \: (T _{1} \cos37^{o} )d - (T _{2} \cos53^{o} )(2 - d) = 0$

$\rm \: T _{1} ( \dfrac{4}{5} )d - ( \dfrac{3}{4} T _{1} ) \ (\dfrac{3}{5} )(2 - d) = 0$

$\rm \: \cancel{ T _{1} }( \dfrac{4}{ \cancel5} )d - ( \dfrac{3}{4} \cancel{ T _{1} )} \ (\dfrac{3}{ \cancel5} )(2 - d) = 0$

we get

$\rm \: 4d = \dfrac{9}{4} \times (2 - d)$

$\rm \: 4d = \dfrac{9}{4} \times 2 - \dfrac{9}{4}$

$\rm \: 4d + \dfrac{9}{4} d = \dfrac{9}{2}$

Taking lcm

$\rm \: \rm \: \dfrac{16d + 9d}{ \not{4}} = \dfrac{9}{ \not2}$

$\rm \: \to \: d = \dfrac{18}{25} m$

Answer 2

Answer:

We can see the bar is in equilibrium so

\rm \: \to \sum F = 0→∑F=0

\rm \to \: \Sigma \tau = 0→Στ=0

Now take 36.9⁰ as 37⁰ and 53.1 as 53⁰ for making easy calculations

=> Horizontal force

\rm \: T_1 \sin37 ^o = T_2 \sin53 {}^{o}T

1

sin37

o

=T

2

sin53

o

Now value of sin37⁰ is 3/5 and sin53⁰ is 4/5

\rm \: T_1 \times \dfrac{3}{5} = T_2 \times \dfrac{4}{5}T

1

×

5

3

=T

2

×

5

4

\boxed{\rm \: T_1 \dfrac{3}{4} = T_2 }

T

1

4

3

=T

2

=> Vertical force

\rm \: T _{1} \cos37^{o} + T _{2} \cos53^{o} = wT

1

cos37

o

+T

2

cos53

o

=w

\rm \: T_1 \times \dfrac{4}{5} + T_2 \times \dfrac{3}{5} = wT

1

×

5

4

+T

2

×

5

3

=w

now we have

\to\rm \: T_1 \dfrac{3}{4} = T_2→T

1

4

3

=T

2

So we get

\rm \: T_1 \times \dfrac{4}{5} + \dfrac{3}{4} T_1 \times \dfrac{3}{5} = wT

1

×

5

4

+

4

3

T

1

×

5

3

=w

\boxed{\rm \: T_1 \dfrac{5}{4} = w}

T

1

4

5

=w

Now we have three unknown value or 2 equation now we using

\Sigma \tau = 0Στ=0

So we know that torques in clockwise is +ve

and torques in anticlockwise is - ve

\rm \: (T _{1} \cos37^{o} )d - (T _{2} \cos53^{o} )(2 - d) = 0(T

1

cos37

o

)d−(T

2

cos53

o

)(2−d)=0

\rm \: T _{1} ( \dfrac{4}{5} )d - ( \dfrac{3}{4} T _{1} ) \ (\dfrac{3}{5} )(2 - d) = 0T

1

(

5

4

)d−(

4

3

T

1

) (

5

3

)(2−d)=0

\rm \: \cancel{ T _{1} }( \dfrac{4}{ \cancel5} )d - ( \dfrac{3}{4} \cancel{ T _{1} )} \ (\dfrac{3}{ \cancel5} )(2 - d) = 0

T

1

(

5

4

)d−(

4

3

T

1

)

(

5

3

)(2−d)=0

we get

\rm \: 4d = \dfrac{9}{4} \times (2 - d)4d=

4

9

×(2−d)

\rm \: 4d = \dfrac{9}{4} \times 2 - \dfrac{9}{4}4d=

4

9

×2−

4

9

\rm \: 4d + \dfrac{9}{4} d = \dfrac{9}{2}4d+

4

9

d=

2

9

Taking lcm

\rm \: \rm \: \dfrac{16d + 9d}{ \not{4}} = \dfrac{9}{ \not2}



4

16d+9d

=



2

9

\rm \: \to \: d = \dfrac{18}{25} m→d=

25

18

m