# Bsby Crib Torque problem

1. Dec 3, 2003

### zeronem

Heres a torque problem that has baffled me.

Soon-Yi is building a mobile to hang over her baby's crib. She hangs a .020-kg toy sailboat .010 m from the left end and a .015-kg toy truck .20 m from the right end of a bar .50 m long. If the lever arm itself has negligible mass, where must the support string be placed so that arm balances?

.077 meters

I just got it out of messing around in Mathematics. Someone help solve this please.

EDIT MESSAGE: I just got .4 meters as my answer. This time i wasn't so careless on my Mathematical work. Please check both my answers.

ALSO,, im using the acceleration due to gravity as 10 m/s^2

Last edited: Dec 3, 2003
2. Dec 3, 2003

### gnome

1. The question asks "where must the support string be placed?" Just giving a number does not answer the question. You should be specifying a distance from one end or the other.

2. Doesn't matter though. They both look wrong to me. How are you getting your results? Show your work.

3. Dec 3, 2003

### zeronem

Ill show my work,, but if its in sloppy manner im sorry.

I should be in bed, but ill take the time to show my work. The reason I didn't show my work is because im fixing to go to sleep, so I figured I didn't have enough time.

The answers are in meters, and basically you are right. I must say how far it is form the right or the left.

(.2 X .010) + (.15 X -.2) + (f X d) = 0

.002 + -.03 + (f X d) = 0

.010 + .2 - .5 = -.21

.002 + -.03 + (f + -.21) = 0

-.028 + (f X -.21) = 0

.028 = -.21f
f = -.1333333333

.002 + -.03 + (-.1333333333 X D) = 0

-.028 + (-.13333333333 X D) = 0
-.1333333333d = .028

d=-.210000001

Look at my answer, its different!

Or, I might need to take -.210000001 and add to .50

I get .3 meters

Perhaps this is from the left or the right?

EDIT MESSAGE: Sorry for any rudeness, but im extremely Fustrated right now because, I can't seem to get the logic behind this. Im so close to killing myself because I feel extremely stupid because I can't even form my own logical way of solving this problem, and I myself wish to be a Mathematician and Theoretical Physicist. How can I become such, when I cant solve this Kindergarden Problem. Mathematics came natural to me, but I cant do one logical Physics Problem.

Last edited: Dec 4, 2003
4. Dec 4, 2003

### Origian

Hi there,

I just did this, and although my answers look right they are not very nice, round answers. Here is a cheap ASCII diagram of the problem:

|
|
-1----2---

EDIT: OK the diagram didn't come out too good....

The 1 is the 0.2kg sailboat, and the 2 is the 0.015kg truck. As the total bar is .5m, the sailboat is 0.01m from one end and the truck is 0.2m from the other, the distance between the two is .5-.2-.01=0.29m. Now, consider the actual forces acting on the two objects - their weight force due to gravity (F=ma, so F = mg) - and the Torque the two objects will exert around the point where the support string. As t=Fd, T = mgd. In order for the system to be 'balanced', the torque for object 1 has to be the same as the torque for object 2 - so T1=T2, or m1 * g * d1 = m2 * g * d2. So:

0.2 * g * d1 = 0.015 * g * d2

But we know that d1 + d2 = 0.29. So d1 = 0.29 - d2. The equations can then be solved simultaneously to give your two answers for the distances from the support string.

This may be totally wrong (I've just finished High School), but hey, I tried.

Origian.

5. Dec 4, 2003

### HallsofIvy

[/quote](.2 X .010) + (.15 X -.2) + (f X d) = 0[/quote]

"0.01 m" and "0.2 m" are measured from the ends. Where are you taking into acount the fact that the bar is .5 m long?

I think what you are doing is asserting that the total torque around some point is 0. That's correct: since the mobile is not to rotate, the torque around ANY point must be 0- but you have to SAY what point!

Let's set the torque around the left end point of the bar equal to 0. The .02 kg sailboat (weight .02g N) is 0.01 m from the left end so its torque is (-.01)(.02g)= -0.0002g N-m (negative because it will tend to "twist" the bar clock-wise around the left end. The 0.015 kg truck (weight 0.015g N) is .2 m from the right end of the bar and so 0.5- 0.2= 0.3 m from the left end. Its torque will be (-0.3)(0.015g)= -0.0045g N-m (negative for the same reason.

Take d to be the distance, from the left endpoint, to the suspension point. It is supporting 0.015+ 0.002= 0.017 kg or 0.017g N weight. The torque about the left end is (0.017g)d N-m and is positive because the upward pull would "twist" the rod counter-clockwise. In order to have no twist, we must have total toque equal to 0: -0.002g- 0.0045g+ 0.017g d= 0 (notice that "g" cancels)
-0.0067+ 0.017d= 0
0.017d= 0.0065 so d= 0.38235 (or 0.38 to 2 sig. figures). The support should be 0.38 meters from the left end of the rod. (By the way: you will really impress your teacher if you write the answer out in a complete sentence like that! It may also help you see if you have completely answered the question.)

Another way to do this is to find the torque around the (unknown) support position. Let d again be the distance from the left end of the rod to the support. Then the distance from the sailboat to the support is d- 0.01 and the torque due to the sailboat is (0.02g)(d- 0.01).
The (signed) distance from the truck to the support is 0.3- d and the torque is (0.015g)(0.3-d).
In order that the rod "balance", these must be equal:
(0.02g)(d- 0.01)= (0.015g)(0.3-d).
0.02d- 0.002= 0.0045- 0.015d
(0.02+ 0.015)d= 0.0045+ 0.002
0.017d= 0.0047 exactly as we had before.

6. Dec 4, 2003

### zeronem

I thought the object .015kg toy being placed at the right of the .5 meter bar would cause it to go ClockWise,, since its weight is applying on the right of the bar. Causing it to move down in a clock Wise motion thus being negative and not positive. As the Object on the left side being .02kg making the bar go counter-clockwise causing it to be positive. I recognize that Trigonometry can be applied in this area. I still dont quite understand. Please explain more clear.

Edit Message: I took the idea that anything being driven down by an object will be negative. I then found that I should see the distance of the object as being from the left instead of the right, so I subtracted .2 - .5 and got -.3

(.02)(10)(-.01) + (.015)(10)(-.3) + (F x D) = 0

The force will equal (.015)(10) + (.02)(10) = .2 + .15 = .35

Now

(.2)(-.01) + (.15)(-.3) + (.35)d = 0

= -.002 + -.045 + (.35)d = 0

= -.047 + (.35)d = 0

.35d = .047

distance is .134285714 m from the left

I believe that is my answer

Now if I were to see the distances of the objects as they are from the right instead of from the left, will I get the same answer? Or will I get a different answer, only the answer is correct for the objects seen from the right instead of the left.

Last edited: Dec 4, 2003