# Help with 2D force problem

1. Oct 27, 2013

### gray_

1. The problem statement, all variables and given/known data
An object is hanging from three ropes. The first rope is vertical and has a tension of 16N. The second rope leads to the right and makes an angle of 70 degrees with the ceiling and also has a tension of 16N. The third rope leads to the left and makes an angle of 40 degrees with the ceiling and has a tension of 7.2N. Find the mass of the hanging object.

2. Relevant equations
t=mg?
f=ma?

2. Oct 27, 2013

### cepheid

Staff Emeritus
Welcome to PF,

You must also provide item 3: The Attempt at a Solution. What have you done so far on this problem? Hint: what, according to Newton's second law, must be true about the vertical components of all the forces acting on the object? How do you find the vertical components of each force vector?

3. Oct 28, 2013

### gray_

Pretty sure I figured this out. I calculated 16sin70+7.2sin40+16 to get a sum of 35.6, which I divided by 9.8 (gravity) to get an answer of 3.63kg=m. I used trigonometric functions (sine, cosine, tangent) to find my answer. It was easy to figure out once I read through the problem carefully, just wanted to make sure I'm doing this right as this problem is used as a basis for the others I'm required to do. Thanks!

4. Oct 28, 2013

### gray_

Sorry in advance if I'm meant to post a new topic for this, I checked and couldn't find a rule on it, but I have another question in relation to the previous one. This time, instead of finding the mass of the object, you're supposed to work backwards to find the tension force. The question:

An object with a mass of 25kg is hanging from two ropes. The first rope leads to the left and makes an angle of 32 degrees with the ceiling and also has a tension of 16N. The second rope leads to the right and makes an angle of 23 degrees with the ceiling. Find the tension force of the second rope.

My solution:
16cos32+FTensin23=m
13.56+FTensin23=25
13.56+FTen0.39=25
13.95+FTen=25
(13.95-13.95)+FTen=25-13.95
FTen=11.05???

The correct answer is FTen=29N. I can't find where I'm going wrong with this. Any help would be appreciated.

5. Oct 28, 2013

### cepheid

Staff Emeritus
1. Why are you using cosine instead of sine when finding the vertical component of the 16 N force?

2. How the hell did you get from here:

13.56+FTen0.39=25

to here?

13.95+FTen=25

6. Oct 28, 2013

### gray_

1. I did use sine...did you mean that the other way around?
2. I added 13.56+0.39. I know this can't be done but I'm asked to show my process of thinking and this is what went through my head. I really can't figure this one out.

Edit: I'm sorry, I did use cosine for that vertical component...that was my mistake, copied it down wrong.

16sin32+FTensin23=m
8.47+FTensin23=25
8.47+FTen0.39=25
And here's where I'm getting stuck

Last edited: Oct 28, 2013
7. Oct 28, 2013

### Staff: Mentor

In your first equation, the cos32 should be sin32, and the right hand side should be the weight mg. After you solve for the tension FT, you need to confirm that this is also consistent with the horizontal force balance.

Chet

8. Oct 28, 2013

### gray_

16sin32+FTensin23=w
8.47+FTensin23=245N
8.86FTen=245N
/8.86
FTen=27.65N

Right?

Last edited: Oct 28, 2013
9. Oct 28, 2013

### cepheid

Staff Emeritus
No, not right. You seem to be having trouble with basic algebra. You can't just add the sin(23°) to the 8.47, because it's being multiplied by something else! You have:

8.47 + (Ft)sin(23°) = mg

You have to solve for Ft, which means isolating it on one side of the equation. So, first, subtract 8.47 from both sides of the equation:

(Ft)sin(23°) = mg - 8.47

Now, in order to get Ft by itself, you have to divide both sides of the equation by sin(23°)

Other tips:

1. Leave your equations in terms of symbols. Don't substitute in any numbers until the end, when you've isolated the quantity of interest on one side of the equation. It's much cleaner, easier to follow, and requires only one use of the calculator instead of several, which reduces error.

2. You can get superscripts and subscripts by using the X2 and X2 buttons above the reply box. Alternatively you can do it manually by using the SUP and SUB tags. E.g. [noparse] Ft [/noparse] produces Ft.

3. It's also pretty weird the way your Ften and sin23 just blended into each other. At least write Ften*sin23 or (Ften)*sin23 to make it clear that it's two distinct things being multiplied by each other.

10. Oct 28, 2013

### gray_

I got it now. I guess all I needed to see was that second step of the algebra...I don't know why I wasn't getting it before. Sorry for bothering you with such a silly mistake. Thanks for the help, both of you!

11. Oct 29, 2013

### Staff: Mentor

Don't thank us so quickly. I also indicated that you need to check the horizontal force balance to make sure that your answer also satisfies that balance. If not, the problem is not well-posed. There is no way that, in this problem, both the horizontal and vertical force balances can be satisfied with the same value of FT. Sorry, it's an ill-posed problem.

Chet

12. Oct 29, 2013

### gray_

I actually ended up asking my teacher about this. She guided me through the solution:
16sin32+FTsin23=mg
8.47+FT(0.39)=25(9.8)
8.86FT=245
FT=245/8.86= Roughly 29N, which was the answer I was supposed to get. And this was the same answer I was getting earlier.

13. Oct 29, 2013

### Staff: Mentor

Sorry, but both you and your teacher need to learn algebra.

14. Oct 29, 2013

### gray_

I was unsure of her answer at first so I also doublechecked with another physics teacher in the building. This is not only the answer they both gave me, but it's the one given in our textbooks. I think at this point I'd be better off trusting the answer I'm getting by three different sources.

15. Oct 29, 2013

### cepheid

Staff Emeritus
Well, that's ridiculous, because the algebra you and your teacher used is obviously wrong, for the reason I cited above. Basically you said that A + FT*B = (A+B)*FT. You can see that that's wrong, can't you? Plug in some random numbers. Let A = 2, FT = 3, and B = 4

2 + 3*4 = 2 + 12 = 14

(2+4)*3 = 6*3 = 18

These two things are NOT the same.

But since you seem unwilling to believe the "the algebra is wrong" argument, here's another. You can easily verify *for yourself* that the answer of 29 N is wrong. Suppose the tension is 29 N (and let's call this rope 1). Then what is the vertical (y) component of that force?

F1y = (29 N)sin(23°) = 11.33 N

F2y = (16 N)sin(32°) = 8.48 N

So, the sum of the two upward forces is 19.81 N.

What is the downward force on the mass? It's its weight: mg = (25 kg)*(-9.81 N/kg) = -245.25 N.

So, the upward force is 19.8 N and the downward force is -245.25 N. Do these two balance (i.e. sum to zero)? NOT EVEN CLOSE.

Conclusion: 29 N is not the right answer. Also, even if you calculate the answer correctly, It does not satisfy the horizontal force balance, as Chestermiller hinted at. So, whoever made this problem screwed up.

Last edited: Oct 29, 2013
16. Oct 29, 2013

### Staff: Mentor

cepheid and I stand by what we said. I feel sorry for you that your teachers and your textbook are giving you such bogus information (and don't know algebra). In any event, you yourself have no excuse for not being able to do such simple algebra correctly. cepheid explained the correct procedure, but you ignored it. To illustrate how wrong you are, try substituting Ft=606.5 into your starting equation and see if it satisfies this equation. A linear algebraic equation can have only one solution.

17. Oct 29, 2013

### gray_

I get what you're saying. I'm not opposed to believing the algebra is wrong. You have to understand that I'm, probably quite obviously, new to physics; I'm a sophomore and was put in AP Physics without having ever taken the class before, and when two different people can show me how they got this answer and I can understand it, it's going to confuse me when I hear a different answer. I get it now though, your explanation makes a lot of sense. I really appreciate you taking the time to explain this to me so thank you.

18. Oct 29, 2013

### Staff: Mentor

Not having had physics before and being a sophomore who was put in AP Physics is no excuse for not being able to do freshman elementary algebra. You certainly must have had that before.

19. Oct 29, 2013

### gray_

I've actually never had an algebra class. I haven't been properly taught the basics of algebra; it's a skill we are expected to learn over time, I guess. I do know how to do it, though, but I will make mistakes. I'm not perfect. The format asks the poster to write out our attempt at the solution, so I wrote what went through my head and I recognized I had it wrong.

20. Oct 29, 2013

### cepheid

Staff Emeritus
Oh ok. That's too bad. In solving high school physics problems, you definitely benefit from knowlege of algebra (not to mention that it ought to have been taught by now regardless IMO -- doesn't sophomore mean second year of high school in your terminology?).

An algebra concept that might help clarify this situation is the distributive property of multiplication. Multiplication is distributive, meaning that when you have something like:

(A + B)*C

I.e. C is being multiplied by a sum or two or more terms, you can "distribute" the C amongst *each* of the terms in the sum. I.e., when you bring it inside the parentheses, you must multiply it separately by each term in the sum. This makes sense: if I have a sum of two things, then multiplying the sum by C is the same as multiplying each thing by C, and then adding the results together. So, this would become

(A + B)*C = AC + BC

Clearly AC + BC is not the same thing as just A + BC. Does that make sense?

In the specific example you were considering, we had A + BC, where A = 8.47, C = FT, and B = 0.39. This is not the same as AC + BC, yet your teacher was saying that it was. Since the C was not present on both terms, you could not factor it out the way he/she did.

I hope that helps a bit.

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