Mass Hanging in Center of Chain Problem

[rocapp]

Homework Statement

A mass of 200 kg is hanging directly in the center of a chain; the chain makes a 20^o angle from its horizontal. The chain will break if more than 2000 N of force are applied at any point on the chain. Will the chain break?

Homework Equations

F=ma
Tfy=mg(sin$\Theta$)
g=10m/s

The Attempt at a Solution

2000N = 2Tf(sin20)
Tf1 = 1000/sin20
Tf1 = 1095 N
Tf2 = 1095 N
Tftotal = 1095N + 1095N
Tftotal = 2190 N
The chain will break.

 

[collinsmark]

Homework Statement

A mass of 200 kg is hanging directly in the center of a chain; the chain makes a 20^o angle from its horizontal. The chain will break if more than 2000 N of force are applied at any point on the chain. Will the chain break?

Homework Equations

F=ma

Newton's second law is fine here. What it means in this case is that since nothing is accelerating, all the forces involved must sum up to 0 (the vector sum).

Tfy=mg(sin$\Theta$)

The above equation is incorrect, as applied to this problem.

g=10m/s

I think you mean units meters per second squared.

The Attempt at a Solution

2000N = 2Tf(sin20)

Okay, that's better! I assume the "2000 N" comes from mg = (200 [kg])(10 [m/s²]) = 2000 N. So far so good.

Tf1 = 1000/sin20

Everything in your attempted solution is fine so far.

Tf1 = 1095 N
Tf2 = 1095 N
Tftotal = 1095N + 1095N

Wait, something's not right here. Where did the 1095 N come from?

And why are you breaking it into 2 parts? There's no need to do that because you've already accounted for the two sections of chain in your "2000N = 2Tf(sin20)" formula.

Tftotal = 2190 N

So that's not quite right.

If it helps things make more sense, take another look at your free body diagram (FBD).

 

[rocapp]

Hey thanks for the response!
The 1095N comes from
1000N/sin20 = Tf
Tf = 1095N
Is Tf the total tension force?

 

[collinsmark]

Hey thanks for the response!
The 1095N comes from
1000N/sin20 = Tf
Tf = 1095N

I think your calculator is set to radians. The problem statement said the angle was 20^o, not 20 radians.

Is Tf the total tension force?

You're the one that defined what Tf stands for. You tell me!

Really though, make sure to draw all the forces out on your FBD. That way, if you forget what one of the variable names are, your FBD will remind you what it is.

(But yes, I was assuming that when you set up your 2000 [N] = 2T_f(sin20) equation, that T_f is the magnitude of the tension on the chain. But you should check your FBD to be sure.)

 

[rocapp]

Attached is my FBD.

You were spot on, I believe.

I used a calculator where I could specify degrees, and I came out with 2923.8 N. In other words, the chain will break.

Weirdest thing is that it was my cell phone's calculator. For what possible reason is that in radians?

 

[collinsmark]

I used a calculator where I could specify degrees, and I came out with 2923.8 N. In other words, the chain will break.

That looks right to me.

Weirdest thing is that it was my cell phone's calculator. For what possible reason is that in radians?

Working in radians make a lot of sense in many situations. But sometimes degrees is more convenient. Whichever cell phone you have (or whichever app you are using on your cellphone if that is the case), there should be a settings parameter that will allow you to specify radians or degrees, before using the trigonometric functions. Look in the cell phone calculator's settings menu or something like that and you should be able to change it back and forth as necessary.

Or you can always convert back and forth manually using,
$$\frac{\mathrm{AngDeg}}{180} = \frac{\mathrm{AngRad}}{\pi}$$

 

[rocapp]

Thank you very much! I believe I understand tension force problems now!

 

[rocapp]

I do have one question, though. How is the magnitude of the total tension force greater than the magnitude of the weight? Where does this force come from? I read that tension force is the opposite of compression force. Is this related? Thanks again!

 

[azizlwl]

I do have one question, though. How is the magnitude of the total tension force greater than the magnitude of the weight? Where does this force come from? I read that tension force is the opposite of compression force. Is this related? Thanks again!

If you pull directly an object up, the force must be equal to weight since total force is applied to lift the object.
If you are not pulling directly, only part(component) doing the job, which opposite to the weight.
The total force is the sum of its parts

 

[Xisune]

Only the vertical component of the chain has the equal half the magnitude of the weight, this vertical component is Tsin20, if you set that equal to (200*9.8)/2 and solve for tension, the tension will be greater.