Tension in two strings with 67 kg mass suspended by them

[GiantSheeps]

Homework Statement

A light string has its ends tied to two walls separated by a distance equal to five-ninths the length of the string as shown in the figure. A 67 kg mass is suspended from the center of the string, applying a tension in the string. What is the tension in the two strings of length L/2 tied to the wall? The acceleration of gravity is 9.8 m/s^2 . Answer in units of N.Here's the diagram, tell me if this link isn't working, I've never used google drive for this before

https://drive.google.com/file/d/0By72ra-d5ponYWZCYllGOW90NFE/view?usp=sharing

Homework Equations

Torque T=rFsinΘ

The Attempt at a Solution

So I know that the two angles must be equal, I know that the tension must be evenly distributed between the two so I know that the tension in the two lengths of string must be equal and I know the downward force on the block mg = 67*9.8 = 656.6, and the upward force on the block must equal the downward force so the string provide an upward force of 656.6

but here is where I'm stuck. I have no idea what to do next and I don't know what to do with the rope length fractions. Any help would be greatly appreciated. Thanks in advance!

 

[Chestermiller]

If you call T the tension in the string, from the geometry, what are the components of the tension in the horizontal and vertical directions? (Think Pythagorean theorem).

Chet

 

[GiantSheeps]

Do you mean like T^2=Thorizontal^2+Tvertical^2 ?

 

[Chestermiller]

Do you mean like T^2=Thorizontal^2+Tvertical^2 ?

No, I mean Tcosθ and Tsinθ. What are the sine and cosine of θ?

Chet

 

[GiantSheeps]

Oh okay so the vertical component would Tsinθ and the horizontal component would Tcosθ

 

[Chestermiller]

Oh okay so the vertical component would Tsinθ and the horizontal component would Tcosθ

Yes. So all you need to do is use the geometry to determine sinθ and cosθ.

Chet

 

[GiantSheeps]

Yes. So all you need to do is use the geometry to determine sinθ and cosθ.

Chet

But how can I do that without knowing the tension in the string or what θ equals?

 

[GiantSheeps]

Yes. So all you need to do is use the geometry to determine sinθ and cosθ.

Chet

I know that Tsinθ and Tcosθ = 358.3, can I use that to find what theta equals?

 

[Chestermiller]

You need to determine θ from the geometry.

 

[GiantSheeps]

You need to determine θ from the geometry.

But how do I do that?

 

[Chestermiller]

But how do I do that?

You have an isosceles triangle, and you know all three sides of the triangle. Try dropping a normal from the apex to the base. Do you see any right triangles when you do this? One of the angles of the right triangle is θ. Can you see what to do now?

Chet

 

[GiantSheeps]

I got two right triangles

each right triangle has a hyoptenuse of 1/2L and one leg is 5/18L

I said cosθ =(5/18) / (1/2) = (5/9)

so cosθ = 5/9L, then I said cos^-1(5/9) = 56.251, meaning θ = 56.251

Did I do that right?

 

[Chestermiller]

I got two right triangles

each right triangle has a hyoptenuse of 1/2L and one leg is 5/18L

I said cosθ =(5/18) / (1/2) = (5/9)

so cosθ = 5/9L, then I said cos^-1(5/9) = 56.251, meaning θ = 56.251

Did I do that right?

Yes, but you didn't actually have to determine the angle. All you need is the sine and cosine of the angle. You already have the cosine. What's the sine?

 

[GiantSheeps]

Yes, but you didn't actually have to determine the angle. All you need is the sine and cosine of the angle. You already have the cosine. What's the sine?

Sine is opposite over hypotenuse, and the hypotenuse is 1/2 L, but I don't know the opposite. The opposite would just be the length of the normal, but I don't know how I would determine that

 

[Chestermiller]

Sine is opposite over hypotenuse, and the hypotenuse is 1/2 L, but I don't know the opposite. The opposite would just be the length of the normal, but I don't know how I would determine that

You know two of the sides of a right triangle, and you don't know how to determine the third side. Is that what you are saying? Pythagoras?

 

[GiantSheeps]

oh of course I can't believe I missed that. So the length of the normal is the square root of 14/81. Now I'd take that and divide it by (5/18) to get 1.496662955, which is the sine. And I have cosθ=5/9 and sinθ=1.496662955. How can I use this information to find the tension in the string?

 

[Chestermiller]

oh of course I can't believe I missed that. So the length of the normal is the square root of 14/81. Now I'd take that and divide it by (5/18) to get 1.496662955, which is the sine. And I have cosθ=5/9 and sinθ=1.496662955. How can I use this information to find the tension in the string?

Check your arithmetic. The sine of an angle can't be greater than 1. I get 0.831 for the sine.

 

[GiantSheeps]

oh yes I accidently divided by the length of the adjacent rather than the length of the hypotenuse. So cosθ=5/9 and sinθ=0.831. How can I use this information to find the tension in the string?

 

[Chestermiller]

oh yes I accidently divided by the length of the adjacent rather than the length of the hypotenuse. So cosθ=5/9 and sinθ=0.831. How can I use this information to find the tension in the string?

If T is the tension in the string, what is the vertical component of the tension (algebraically)?

 

[GiantSheeps]

T = vertical component + horizontal component, so the vertical component would = T - the horizontal component?

 

[Chestermiller]

T = vertical component + horizontal component, so the vertical component would = T - the horizontal component?

Tsinθ

Have you drawn a free body diagram showing the forces acting on the point from which the weight is hung?

Chet

 

[GiantSheeps]

Oh thank you so much! I just realized what I had to do divide the downward force (656.6N) by the vertical component times 2! Thank you so much for your help it is greatly appreciated. Just out of curiosity, did you know exactly how to do this problem the instant you saw it, or did you have to work it out a bit? You do seem like some sort of expert at this sort of thing

 

[Chestermiller]

Oh thank you so much! I just realized what I had to do divide the downward force (656.6N) by the vertical component times 2! Thank you so much for your help it is greatly appreciated. Just out of curiosity, did you know exactly how to do this problem the instant you saw it, or did you have to work it out a bit? You do seem like some sort of expert at this sort of thing

This isn't about me. It's about you. You are assigned homework problems so that, in addition to the fundamentals you learned in class, you can develop proper technique to attack problems. Here are some techniques that are critical:

1. Before starting to think about forces and equilibrium, always first analyze the geometry of the system. In problems involving 1D, this is trivial, but in 2D and 3D, it is very important. In problems involving motion, analyzing the geometry also includes analyzing the kinematics of the motion.

2. In real estate, the three most important things is LOCATION, LOCATION, LOCATION. In mechanics, the three most important things are FREE BODY DIAGRAMS, FREE BODY DIAGRAMS, FREE BODY DIAGRAMS. It is not possible to overemphasize the importance of drawing and applying free body diagrams in solving mechanics problems.

Chet

 

[dean barry]

The force acting down due to the mass (m*g) can be split into two, then deal with one side only, imagine this force as the vertical leg of the right angle triangle.
A key point with this excercise is that as the angle diminishes, the Value of T rises dramatically, as the angle tends to 0, the tension reaches infinity.