Question about the stretch of rope

[aditya ver.2.0]

A rope of length L₁ is stretched from both ends by A and B ( both are exerting equal amount of force F₁ and F₂, where F₁=F₂)to keep it rigid. Just then C ties a metallic block of mass M₁ at the center of the rope.Due to the weight of the block, the rope hangs down and creates a depression of D₁ meters.

Q.1 At what rate (in Newton) should A and B pull the string so that the depression starts to reduce by H₁ m/sec, in the presence of the metallic block?

Q.2 At what rate (or differential rate) should both the ends of the rope move, so that the depression reduces by H₁ m/sec?

 

[sophiecentaur]

Hi. This looks like a Homework / Coursework question and it should probably be in that forum.
As we don't just answer such questions 'for free' (PF is far too mean to spoon feed people!), we need to see what you, yourself, have done towards answering the question.
For instance, what are the forces acting on the Mass M1 and what would be the tension in the rope?

 

[aditya ver.2.0]

Sorry Sir, I am beginner in this section. Urgently want to learn how to solve such question. Please...

 

[Drakkith]

Sophie is correct. This question belongs in the homework forums. Please repost it there and use the template provided when you click on New Post. Note that PF is not here to teach you how to solve such questions from scratch. If you have no idea where to begin then you should seek help from your teacher or someone in your class.

 

[pmacias]

I would approach this question by first considering the forces acting on the rope. Since the rope is being stretched from both ends by equal forces, it is in a state of equilibrium and not moving. When the metallic block is added, its weight creates a downward force on the rope, causing it to hang down and create a depression.

To answer the first question, we need to determine the force required to counteract the downward force of the block and reduce the depression by H1 meters per second. This can be done by using the equation F=ma, where F is the force, m is the mass of the block, and a is the acceleration. Since we want the depression to reduce by H1 meters per second, the acceleration of the block must be equal to H1. Therefore, the force required to achieve this acceleration would be F=M1*H1. Since A and B are exerting equal forces, each of them would need to pull with a force of F/2, or (M1*H1)/2, in order to achieve the desired reduction in depression.

For the second question, we need to determine the rate at which both ends of the rope should move in order to reduce the depression by H1 meters per second. This can be done by using the equation v=√(2gh), where v is the velocity, g is the acceleration due to gravity, and h is the height of the depression. In this case, the height of the depression is D1, so the velocity required to reduce it by H1 meters per second would be √(2g(D1-H1)). Since both ends of the rope are moving, each end would need to move at a rate of v/2, or √(2g(D1-H1))/2, in order to achieve the desired reduction in depression.

In conclusion, in order to reduce the depression by H1 meters per second, A and B would need to pull the rope with a force of (M1*H1)/2 and both ends of the rope would need to move at a rate of √(2g(D1-H1))/2. It is important to note that these calculations assume ideal conditions and may vary in a real-world scenario.