Physics, asked by Anonymous, 11 months ago

\huge\boxed{\fcolorbox{blue}{orange}{Answer question no. 44}}

Attachments:

Answers

Answered by nirman95
16

Answer:

Given:

A swing having 20 kg mass. It is held by 2 ropes on either side . 2 persons are sitting on the swing as shown in the attached photo.

To find:

Value of T1 and T2

Calculation:

As per Translational equilibrium , we can say that :

 \large{ \sf{ \Sigma(forces) = 0}}

 \large{ \sf{ =  > T1 + T2 - 50g - 20g - 30g = 0}}

 \large{ \sf{ =  > T1 + T2  = 50g  + 20g + 30g }}

As per rotational Equilibrium :( considering the axis to be the point where 1st rope is attached)

 \large{ \sf{ \Sigma(torques) = 0}}

   \large{ \sf{=  > (50g \times 20) + (20g \times 50)}}  \\  \large{ \sf{+ (30g \times 80)  - (T2 \times 100) = 0}}

  \large{ \sf{=  > T2 = 44g = 440  \: N }}

Putting value of T2 into 1st equation, we get :

 \large{ \sf{ =  > T1 + 44g = 50g  + 20g + 30g }}

  \large{ \sf{=  > T2 = 56g = 560  \: N}}

Attachments:
Answered by Feirxefett
15

Answer:

Given:

A swing having 20 kg mass. It is held by 2 ropes on either side . 2 persons are sitting on the swing as shown in the attached photo.

To find:

Value of T1 and T2

Calculation:

As per Translational equilibrium , we can say that :

\large{ \sf{ \Sigma(forces) = 0}}Σ(forces)=0

\large{ \sf{ = > T1 + T2 - 50g - 20g - 30g = 0}}=>T1+T2−50g−20g−30g=0

\large{ \sf{ = > T1 + T2 = 50g + 20g + 30g }}=>T1+T2=50g+20g+30g

As per rotational Equilibrium considering the axis to be the point where 1st rope is attached)

\large{ \sf{ \Sigma(torques) = 0}}Σ(torques)=0

\begin{gathered} \large{ \sf{= > (50g \times 20) + (20g \times 50)}} \\ \large{ \sf{+ (30g \times 80) - (T2 \times 100) = 0}}\end{gathered}

=>(50g×20)+(20g×50)

+(30g×80)−(T2×100)=0

\large{ \sf{= > T2 = 44g = 440 \: N }}=>T2=44g=440N

Putting value of T2 into 1st equation, we get :

\large{ \sf{ = > T1 + 44g = 50g + 20g + 30g }}=>T1+44g=50g+20g+30g

\large{ \sf{= > T2 = 56g = 560 \: N}}=>T2=56g=560N

Attachments:
Similar questions