Three masses and Two pulley

[kaienx]

Homework Statement

I have 3 masses with 2 pulleys which is in static equilibrium.
M₁ & M₃= 50.8kg
M₂ = 90kg
Gravity that I used = 10m/s²
Part 1
Prove that the system is in static equillibrium

Part 2
If I were to add another 10kg of mass to M₃ is the force exerted from M₃ the same force as the tension from the rope. Assuming the pulley and rope is massless?
And how do I find the displacement of all the masses (M₁, M₂, M₃)?
Will the system be in an equilibrium or not?
If it's not, show how do you find the acceleration, velocity and displacement for all three masses.

Homework Equations

F = ma

The Attempt at a Solution

For part 1, from the equation I have drawn a FBD for M₂ and have calculated the net forces of F(x) and F(y) for M₂ which is:
Fx = 508 * cos (62.5) - 508 * cos (62.5) = 0;
Fy = (508 * sin (62.5) + 508 * sin (62.5)) - 900 = 0 (I round it off due to calculation approximation);

Am I doing the part 1 correctly? As I am assuming that the forces from M₁ & M₃) is acting on M₂ so I took the forces from there and apply it on the FBD for M₂.

For part 2, I can't really continue if I'm not sure whether part 1 is right (correct me if I'm wrong whether can I still continue on part 2). Even then, I'm trying to find the relationship between F = ma and the a, v, and s(displacement) cause I have been searching for the equation but I seem to be more lost than ever.

Thanks once again and I hope I don't sound too dumb as I am new to statics.

 

[stockzahn]

Part 1 seems to be correct.

For part 2: Think about what is the necessary condition for a second equilibrium after increasing the mass M₃ (hint: the angles α and β will be different then).

 

[kaienx]

For part 2, the necessary condition for an equilibrium on the system is that all net forces must equal to 0, however, when I calculated after adding 10kg on M₃, wouldn't the Fx and Fy for M₂ also changes (correct me if I'm wrong), which means M₂ will travel along the rope of M₃, which will pull M₁ up, resulting in an infinite acceleration thus can't be in an equilibrium? So sorry if I sound dumb or stupid, am really new at this and have been self taught from youtube tutorials and etc.

 

[haruspex]

I'm not sure what is being asked in part 2. Are you to find a position of the masses which is at equilibrium (or show there is none such), or find the instantaneous acceleration from the given position?

 

[kaienx]

I'm not sure what is being asked in part 2. Are you to find a position of the masses which is at equilibrium (or show there is none such), or find the instantaneous acceleration from the given position?

Sorry for not being clear, it is to find a position of the masses after the addition of 10kg at M₃ at equilibrium. However I am just curious of the instantaneous acceleration that will occur during the addition of 10kg mass (thus the displacement/velocity/acceleration). That's why I am kind of stuck here (this question was given by my lecturer to explore actually. :)

 

[haruspex]

Sorry for not being clear, it is to find a position of the masses after the addition of 10kg at M₃ at equilibrium. However I am just curious of the instantaneous acceleration that will occur during the addition of 10kg mass (thus the displacement/velocity/acceleration). That's why I am kind of stuck here (this question was given by my lecturer to explore actually. :)

Ok. It was this line that threw me:

If it's not, show how do you find the acceleration, velocity and displacement for all three masses.

The acceleration etc. will depend on the current position. If there's no new equilibrium, there's no position defined. And if you are going to consider accelerations etc. based on the staring point then it won't stop moving when at the equilibrium point.

Anyway, let's first try to find a new equlibrium. Have you drawn a free body diagram? Define some unknowns that describe the arrangement and the forces in it. What equations can you write down?

 

[stockzahn]

For part 2, the necessary condition for an equilibrium on the system is that all net forces must equal to 0, however, when I calculated after adding 10kg on M₃, wouldn't the Fx and Fy for M₂ also changes (correct me if I'm wrong), [...}

Correct

[...] which means M₂ will travel along the rope of M₃, which will pull M₁ up, resulting in an infinite acceleration thus can't be in an equilibrium?

That's true and as haruspex wrote this only applies for the first moment - then the direction of the net force changes.

So sorry if I sound dumb or stupid, am really new at this and have been self taught from youtube tutorials and etc.

Don't be to strict with yourself - it needs a little bit of training to solve this kind of problems. For the second equilibrium: Start to find the condition in horizontal direction.

 

[Chestermiller]

To move to the new equilibrium, which way do you think the 90 kg mass has to move? Would the other masses move up or down? How would that affect the geometry?

Chet

 

[kaienx]

To move to the new equilibrium, which way do you think the 90 kg mass has to move? Would the other masses move up or down? How would that affect the geometry?
Chet

I can only imagine that the affected mass (M₃) will go down, thus M₂ will go north east due to the resultant force applied from M₃, which then M₁ will have to move up to accommodate to the movement of M₂.

Here's what I got so far:

Fx = 608 * cos (62.5) - 508 * cos (62.5) = 234.6 (rounded up);
Fy = (608 * sin (62.5) + 508 * sin (62.5)) - 900 = 89.9 (I round it off due to calculation approximation);
Thus, the Resultant force is: R = √Fx² + Fy² = 251.2;
and the resultant angle is: Rangle = arctan(89.9 / 234.6) = 21;

I'm not sure whether I'm doing it correctly but then again, after finding this, is the system at equilibrium when the mass with 90kg moves over to its new position?

 

[Chestermiller]

The angles will change also.

 

[kaienx]

The angles will change also.

Meaning? Which angles? is it the α and β? I think my train of thought is stuck as I really do not know how to find the new α and β. So my calculations was wrong? No wonder!

 

[Chestermiller]

Meaning? Which angles? is it the α and β? I think my train of thought is stuck as I really do not know how to find the new α and β. So my calculations was wrong? No wonder!

Yes. α and β both change. You need to find the new α and β. You do this by writing the force balances in the x and y directions in terms of α and β (algebraically), and solving for the new α and β.

 

[insightful]

For part 2, the necessary condition for an equilibrium on the system is that all net forces must equal to 0, however, when I calculated after adding 10kg on M3, wouldn't the Fx and Fy for M2 also changes (correct me if I'm wrong), which means M2 will travel along the rope of M3, which will pull M1 up, resulting in an infinite acceleration thus can't be in an equilibrium?

No, the new unbalanced force will cause a finite acceleration. First, calculate the new equilibrium angles (at least show the equations; solving them is tricky).

 

[kaienx]

No, the new unbalanced force will cause a finite acceleration. First, calculate the new equilibrium angles (at least show the equations; solving them is tricky).

So it will be something like,
Fx = 608 * cos (α) - 508 * cos (β) = 0;
Fy = (608 * sin (α) + 508 * sin (β)) - 900 = 0

I then used algebra to solve for Fx and got
(α) = cos^-1(508 cos (β) / 608) = cos^-1(0.84 cos (β)

This part I am stuck as it got a little tricky honestly.
does cos^-1 cancel out cos (β)?

 

[azizlwl]

Why not use Cosine Law.

 

[Chestermiller]

So it will be something like,
Fx = 608 * cos (α) - 508 * cos (β) = 0;
Fy = (608 * sin (α) + 508 * sin (β)) - 900 = 0

I then used algebra to solve for Fx and got
(α) = cos^-1(508 cos (β) / 608) = cos^-1(0.84 cos (β)

This part I am stuck as it got a little tricky honestly.
does cos^-1 cancel out cos (β)?

Solving these two simultaneous equations for the two angles is a little challenging mathematically, but doable. If you do it by brute force, though, it will be pretty difficult.

 

[insightful]

So it will be something like,
Fx = 608 * cos (α) - 508 * cos (β) = 0;
Fy = (608 * sin (α) + 508 * sin (β)) - 900 = 0

I have M1 and alpha on the left, and M3 and beta on the right on your diagram.
In my equations, alpha and beta are reversed from yours(?).
Anyway, I just used trial and error and got 3 significant figures in 7 trials.

 

[kaienx]

I have M1 and alpha on the left, and M3 and beta on the right on your diagram.
In my equations, alpha and beta are reversed from yours(?).
Anyway, I just used trial and error and got 3 significant figures in 7 trials.

Yes apologies it is supposed to be
Fx = 508 * cos (α) - 608 * cos (β) = 0;
Fy = (508 * sin (α) + 608 * sin (β)) - 900 = 0

(α) = cos-1(608 cos (β) / 508) = cos-1(1.2 cos (β))
replace (a) to Fy's equation I get:
508 * sin(cos-1(1.2 cos (β))) + 608 * sin (β)) - 900 = 0;
508 * sin(cos-1(1.2 cos (β))) + 608 * sin (β)) = 900;

I wanted to factorise but it doesn't seem that there's any common values.
Anyway,
@azizlwl cosine law? as far as I can remember I think need to know the three sides which is currently unknown at the moment, no?

 

[Chestermiller]

$$\cosα=\frac{508}{608}\cosβ$$
$$\sinα=\sqrt{1-\left(\frac{508}{608}\right)^2\cos^2β}=\sqrt{1-\left(\frac{508}{608}\right)^2+\left(\frac{508}{608}\right)^2\sin^2β}$$
Substitution in Fy equation yields:
$$\sqrt{608^2-508^2+508^2\sin^2β}+508\sinβ=900\tag{1}$$If we multiply numerator and denominator of the left hand side of this equation by$\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ$, we obtain:
$$\frac{608^2-508^2}{\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ}=900\tag{2}$$
If we invert Eqn. 2, we obtain:
$$\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ=124\tag{3}$$
If we subtract Eqn. 3 from Eqn. 1, we obtain:
$$1016\sinβ=776$$or$$\sinβ=\frac{776}{1016}$$So, β=49.8 degrees

Chet

 

[azizlwl]

Yes apologies it is supposed to be
Fx = 508 * cos (α) - 608 * cos (β) = 0;
Fy = (508 * sin (α) + 608 * sin (β)) - 900 = 0

(α) = cos-1(608 cos (β) / 508) = cos-1(1.2 cos (β))
replace (a) to Fy's equation I get:
508 * sin(cos-1(1.2 cos (β))) + 608 * sin (β)) - 900 = 0;
508 * sin(cos-1(1.2 cos (β))) + 608 * sin (β)) = 900;

I wanted to factorise but it doesn't seem that there's any common values.
Anyway,
@azizlwl cosine law? as far as I can remember I think need to know the three sides which is currently unknown at the moment, no?

You have 3 sides(vectors) 900, 608 and 508 in equilibrium.

 

[kaienx]

$$\cosα=\frac{508}{608}\cosβ$$
$$\sinα=\sqrt{1-\left(\frac{508}{608}\right)^2\cos^2β}=\sqrt{1-\left(\frac{508}{608}\right)^2+\left(\frac{508}{608}\right)^2\sin^2β}$$
Substitution in Fy equation yields:
$$\sqrt{608^2-508^2+508^2\sin^2β}+508\sinβ=900\tag{1}$$If we multiply numerator and denominator of the left hand side of this equation by$\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ$, we obtain:
$$\frac{608^2-508^2}{\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ}=900\tag{2}$$
If we invert Eqn. 2, we obtain:
$$\sqrt{608^2-508^2+508^2\sin^2β}-508\sinβ=124\tag{3}$$
If we subtract Eqn. 3 from Eqn. 1, we obtain:
$$1016\sinβ=776$$or$$\sinβ=\frac{776}{1016}$$So, β=49.8 degrees

Chet

Thank you! Took me a whole day to understand how did you obtain those values and I think I got the accurate answers using different masses on M₃. So the new position of the M₂ is determined from the new tensions from M₃, am I right?

 

[Chestermiller]

Thank you! Took me a whole day to understand how did you obtain those values and I think I got the accurate answers using different masses on M₃. So the new position of the M_{2 is determined from the new tensions from M3, am I right?}

Sure.

 

[insightful]

So, β=49.8 degrees

Shouldn't this be alpha for the problem as in the diagram?

 

[Chestermiller]

Shouldn't this be alpha for the problem as in the diagram?

I really am unable to clearly make out what's written on the diagram. So, probably yes. All I did was solve the equations that were presented.

 

[kaienx]

@insightful yes that is correct, it should be alpha, apologies as I myself got confused with the whole question. But thanks for pointing that out! :) Anyway, there is no way to find out the displacement from the first equilibrium position to the second equilibrium position with just those two equations?

 

[Chestermiller]

@insightful yes that is correct, it should be alpha, apologies as I myself got confused with the whole question. But thanks for pointing that out! :) Anyway, there is no way to find out the displacement from the first equilibrium position to the second equilibrium position with just those two equations?

No. You need to know the distance between the pulleys (or some characteristic distance).

 

[kaienx]

The distance between pulleys meaning, the red line? (See picture for reference). If I were to have the values, can I assume that I will be able to solve using basic trigonometric formulas seeing that I have the angles needed? (TOA CAH SOH).

 

[Chestermiller]

The distance between pulleys meaning, the red line? (See picture for reference). If I were to have the values, can I assume that I will be able to solve using basic trigonometric formulas seeing that I have the angles needed? (TOA CAH SOH).

Yes.