- #1

Juanda

- 298

- 101

- Homework Statement
- 3 masses are hanging from 2 pullies as shown in the picture.

Assume ##l_1## and ##l_4## to be sufficiently large to allow for equilibrium to be reached if possible.

There is no friction. The pullies are small enough to be considered a point. Constant gravity is at play.

1) Knowing ##m_1##, ##m_2## and ##m_3## find ##\alpha##, ##\beta##, ##l_2## and ##l_3##.

2) If ##m_1## and ##m_3## are known, find the limiting values of ##m_2## that make equilibrium possible. Apply "question 1" to those 2 limiting scenarios.

- Relevant Equations
- Newton.

Vector summation.

If this was just 1 pulley and two masses, then equilibrium is only possible if both masses have the same weight. If we have something as shown, then the possibilities grow. If ##m_2## is too light then it's just as the version with only 2 masses and 1 pulley. On the other hand, if it's too heavy then it'll drag both with it. There is a whole region in the middle of those two scenarios that make equilibrium possible where ##m_2## will hang lower or higher depending on how big it is compared with ##m_1## and ##m_3##.

I could find the limits numerically using a graphic calculator but I'd like to know if it's possible to find them analytically and get more insight about the origin of the solution. If possible, while using just Newton. I'm sure it must be possible to set up the problem so the potential energy is to be minimized and I'd like to solve it that way too eventually but not at the moment.

I'll start with question 1. To find that, it's necessary to apply equilibrium to ##m_2##.

$$\sum F_x=0 \rightarrow -m_1g\sin\alpha + m_3gsin\beta = 0\rightarrow m_1\sin\alpha = m_3\sin\beta \tag{1}$$

$$\sum F_y=0 \rightarrow m_1g\cos\alpha -m_2g+ m_3gcos\beta = 0\rightarrow m_1\cos\alpha + m_3\cos\beta = m_2 \tag{2}$$

From equations ##1## and ##2## it is already possible to find ##\alpha## and ##\beta##. It's not pretty because it's a trigonometric system of equations so I couldn't completely clear the variables. This is what I got.

From ##1##.

$$\alpha=\arcsin(\frac{m_3}{m_1}\sin\beta) \tag{3}$$

Plugging that into ##2## results in.

$$m_1\cos(\arcsin(\frac{m_3}{m_1}\sin\beta) )+ m_3\cos\beta = m_2 \tag{4}$$

I believe ##\beta## cannot be cleared from ##4## so it's only possible to find numerical solutions. Still, from the numerical solution of ##4##, it's possible to plug in the value in ##3## and then we'd have both ##\alpha## and ##\beta## solved. Once that is known, it's now possible to find the values of ##l_2## and ##l_3##.

The sum of the purple vectors must be 0. This gives two equations. Only ##l_2## and ##l_3## are unknown so it's solvable.

$$\sum x=0 \rightarrow l_2 \cos(3\pi/2+\alpha)+l_3\cos(\pi/2-\beta)-b=0 \tag{5}$$

$$\sum y=0 \rightarrow l_2 \sin(3\pi/2+\alpha)+l_3\sin(\pi/2-\beta)-a=0 \tag{6}$$

With that, the first question should be completely solved with the inconvenience of not being able to clear eqn ##4## to get a nice expression for the angles.

Now for question 2, I could only find the limiting values using a graphic calculator and I'd like to know if there is a better method that'd give some insight. I considered eqn ##4## to be a function. To deal with the online graphic calculator I changed ##\beta## with ##x##.

$$m_1\cos(\arcsin(\frac{m_3}{m_1}\sin\beta) )+ m_3\cos\beta = m_2$$

$$\downarrow$$

$$f(x)=m_1\cos(\arcsin(\frac{m_3}{m_1}\sin x) )+ m_3\cos x - m_2 \tag{7}$$

Fixing ##m_1## and ##m_3## it's possible to change the value of ##m_2## and see if there's any ##x## (##x=\beta##) for which ##f(x)=0##. As a first shot, the studied region will be within ##0\leq\beta\leq \arctan(a/b)##. The angle will be 0 when ##m_2## is too heavy and the angle will be parallel to the line between the supports if ##m_2=0##. In the following link, you can see the function.

https://www.desmos.com/calculator/8ddibiq3v0

Lowest possible ##m_2## that makes the function ##f(x)=0## within the shown range.

Greatest possible ##m_2## that makes the function ##f(x)=0## within the shown range.

The limit case where ##m_2## is too heavy is somewhat intuitive. From playing with the function it seems that if ##m_2 = m_1+m_3## then the angles need to be 0 to support it. This can be seen in eqn ##2## although it feels weird both angles approach 0 "at the same speed" as ##m_2## goes lower and lower.

For the smallest ##m_2## though... I have no intuition about where the limiting value comes from. Let me know if you have other point of view or resolution method that could make this a little clearer.