Find Mass of Rope Given Horizontal Force of 20N

[utkarshakash]

Homework Statement

One end of a rope is fixed to a vertical wall and the other end pulled by a horizontal force of 20N. The shape of the flexible rope is shown in the figure. Find its mass.

Homework Equations

http://www.luiseduardo.com.br/mechanics/static/staticproblems_arquivos/image017.jpg

The Attempt at a Solution

Let the wall exert a force F in the tangential direction at the point where rope is fixed. Let the tangent make an angle θ with vertical.
Fcosθ=mg
Fsinθ=20

cotθ= mg/20

But θ is still unknown!

 

[projjal]

But θ is still unknown!

Without θ question is incomplete. The horizontal force applied is 20N : only this information cannot determine the mass of the rope.

 

[SteamKing]

Without θ question is incomplete. The horizontal force applied is 20N : only this information cannot determine the mass of the rope.

Not so fast! The OP hasn't even begun to analyze the rope.

This is a well-known problem in physics, analyzing suspended ropes and chains. There is a solution, and the answer might surprise you.

http://en.wikipedia.org/wiki/Catenary

 

[projjal]

Well, sorry for my incomprehensibility.Its first time I came across such problem.

 

[utkarshakash]

Not so fast! The OP hasn't even begun to analyze the rope.

This is a well-known problem in physics, analyzing suspended ropes and chains. There is a solution, and the answer might surprise you.

http://en.wikipedia.org/wiki/Catenary

Hmm, the article is difficult to understand. Can you please guide me step-by-step on how to solve this problem? I think finding the equation for the curve should be my first step. But I don't have any idea how to do that.

 

[SteamKing]

Hmm, the article is difficult to understand. Can you please guide me step-by-step on how to solve this problem? I think finding the equation for the curve should be my first step. But I don't have any idea how to do that.

You don't have enough information to do all the detailed calculations. However, take a look at the results obtained in the 'Analysis' section of the Wiki article. There's your answer (with a helpful diagram showing all the forces acting on the rope.

Let c be the lowest point on the chain, called the vertex of the catenary, [42] and measure the parameter s from c. Assume r is to the right of c since the other case is implied by symmetry. The forces acting on the section of the chain from c to r are the tension of the chain at c, the tension of the chain at r, and the weight of the chain. The tension at c is tangent to the curve at c and is therefore horizontal, and it pulls the section to the left so it may be written (−T0, 0) where T0 is the magnitude of the force. The tension at r is parallel to the curve at r and pulls the section to the right, so it may be written Tu=(Tcos φ, Tsin φ), where T is the magnitude of the force and φ is the angle between the curve at r and the x-axis (see tangential angle). Finally, the weight of the chain is represented by (0, −λgs) where λ is the mass per unit length, g is the acceleration of gravity and s is the length of chain between c and r.

 

[PhanthomJay]

The figure or statement seems to be missing information. I can take a piece of thin rope and pull on it horizontally with a force of 20 N, and then I can take a piece of heavier rope and pull on it horizonatlly with the same force. In the latter case, the sag will be greater and the angle sharper, but with that info missing, I cannot calculate the mass or weight of the thread.