Answers
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 :
As per rotational Equilibrium :( considering the axis to be the point where 1st rope is attached)
Putting value of T2 into 1st equation, we get :
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