Why Do My Pulley System Calculations Differ From My Teacher's Measurements?

[MironeDagains]

Homework Statement

There's a 200g piece of weight hanging by a string. That string is then split into two strings, each at θ from the horizontal and Φ from the horizontal. These two strings are each connected to force sensors which measure their tension in Newtons. I know how to do this problem, as it's a simple trigonometric or pythagoras problem. The problem I'm having, is the fact that my answers are completely different to what my teacher got. She actually used real sensors and wrote down what she saw when she tested each θ and Φ.
Maybe I don't know what I'm talking about, or maybe my teacher's force sensor is faulty. You decide.
Here's a diagram: http://i.imgur.com/439esTg.png

Homework Equations

F=mg
sin(θ)x0.2kgx9.8=N_F1
sin(Φ)x0.2kgx9.8=N_F2
Triangles, sin cos tan

The Attempt at a Solution

F₁: sin θ x 0.2kg x 9.8
F₂: sin Φ x 0.2kg x 9.8
Here's an image of my method. Please let me know whether it's correct or not. http://i.imgur.com/LTbo8Q9.jpg I cannot reproduce them as words here, as they are almost all diagrams.
Note: The columns "Force F₁ measured", "Angle θ", "Force F₂ measured" and "Angle Φ" were already pre-filled when I was handed this worksheet. The only things I filled in were the "Force F₁ calculated" and "Force F₂ calculated". You will notice how incredibly 'off' my calculated measurements are from her measured measurements.
So, am I going insane? Or does my teacher need to invest in a new force sensor?
Thank you

 

[CWatters]

F1: sin θ x 0.2kg x 9.8
F2: sin Φ x 0.2kg x 9.8

Those aren't correct. What would happen if you make both strings vertical so that θ = Φ = 90? Try it and you will see it's wrong.

The 200g mass is stationary (not accelerating) so what does that say about the net horizontal and vertical forces acting on it?

 

[MironeDagains]

Those aren't correct. What would happen if you make both strings vertical so that θ = Φ = 90? Try it and you will see it's wrong.

But that's exactly the steps taken to solve 2 very similar (practically identical) problems in my physics book?? Look: http://i.imgur.com/nqK5TTY.png what am I missing here?

The 200g mass is stationary (not accelerating) so what does that say about the net horizontal and vertical forces acting on it?

The sum of horizontal forces are equal to zero, and the sum of the vertical forces equal zero. Right?

 

[insightful]

In the sample problems, the weight was divided equally between the two strings. Why is that not the case in this problem?

 

[MironeDagains]

Edit: nevermind. There is no time to wait for a reply. Admins, please delete this thread.

 

[SammyS]

What you are calling F₁ is actually the component of the weight that's in the direction of string 1. Similar for F₂ and string 2.

What you actually need is the vertical component of F₁ added to the vertical component of F₂ to equal the weight.

And as you stated earlier you need the following to be true.

The sum of horizontal forces are equal to zero,

That is the horizontal component of F₁ needs to be equal in magnitude to the horizontal component of F₂ .

 

[haruspex]

Edit: nevermind. There is no time to wait for a reply. Admins, please delete this thread.

The thread may still have value for others, if only in showing how careful you have to be in assuming the same equations for set-ups that are not identical.
Consider the junction between the three strings. There are three forces acting on this point. Apply the usual statics equation in each of the horizontal and vertical directions, $\Sigma F = 0$.

 

[SammyS]

By the way, your teacher's force sensors look to be working pretty well.

 

[MironeDagains]

F₁ = $\frac{1.96cos(θ)}{cos(θ)sin(Φ)+sin(θ)cos(Φ)}$N

and

F₂ = $\frac{1.96cos(Φ)}{cos(θ)sin(Φ)+sin(θ)cos(Φ)}$N

Correct?
​

My answers are much closer to my teacher's now. Which means it is correct.

 

[haruspex]

F₁ = $\frac{1.96cos(θ)}{cos(θ)sin(Φ)+sin(θ)cos(Φ)}$N

and

F₂ = $\frac{1.96cos(Φ)}{cos(θ)sin(Φ)+sin(θ)cos(Φ)}$N

Correct?
​

My answers are much closer to my teacher's now. Which means it is correct.

Yes, those look right. You can simplify the denominator a bit.

 

[insightful]

F₁ = $\frac{1.96cos(θ)}{cos(θ)sin(Φ)+sin(θ)cos(Φ)}$N

and

F₂ = $\frac{1.96cos(Φ)}{cos(θ)sin(Φ)+sin(θ)cos(Φ)}$N

Correct?

​

My answers are much closer to my teacher's now. Which means it is correct.

Are F1 and F2 reversed from your worksheet?

 

[haruspex]

Are F1 and F2 reversed from your worksheet?

Good catch.

 

[MironeDagains]

Are F1 and F2 reversed from your worksheet?

Reversed? Well, F₁ is pointing to the left and F₂ is pointing to the right. As shown in the diagram.

 

[insightful]

Reversed? Well, F₁ is pointing to the left and F₂ is pointing to the right. As shown in the diagram.

Okay, then phi and theta are reversed in your numerators. Re-do a calculation using your above formulas.

 

[MironeDagains]

Okay, then phi and theta are reversed in your numerators.

Yes, they are.

Re-do a calculation using your above formulas.

What? What's that supposed to mean? I already did the calculations, using my new equations, and got answers that were very close to my teacher's. What are you telling me to change and 're-do'?

 

[insightful]

Yes, they are.
What? What's that supposed to mean? I already did the calculations, using my new equations, and got answers that were very close to my teacher's. What are you telling me to change and 're-do'?

When I use your equations, I get the results for F1 and F2 reversed. When I switch phi and theta in the numerators in your equations, I get the correct answers.