Asymmetrical Static Equilibrium

[Mike94]

Homework Statement

A cargo of 200 kg will be free to move vertically along his guide. Two counterweights identical of mass M will lift the cargo. There is a static friction
20 N between the cargo and his guide and a dynamic friction of 5 N. We will neglect all other friction as well as the radius of the pulleys.

Relevant Equations

a) The rope from the cargo to the pulley does what angle at the beginning?

b) What must the masses M be at least to lift the cargo ?

Now, we place mass counterweights M = 400 kg each.


c) What is the acceleration of the cargo a moment after it is lifted (at all departure)? Warning, the acceleration of the rope is not the vertical acceleration of the cargo!


d) During its movement, does the cargo undergo a constant acceleration? (Justify)


e) At what height (of the ground) will it be when its acceleration is zero? (Indices: if a vertical = 0 then ΣFy =?)

a) Answer that I found: 56.31 degrees
b) Answer that I found: 121.71 Kg
For b) not sure if I need to add the 20 N and 5 N to the total of the cargo mass.

 

[kuruman]

(a) and (b) are correct. What are your thoughts about the rest of the problem?

 

[Mike94]

Thank you for the reply !

I was thinking about this for c)

∑Fx max = 1179 N * sin(56.31) = 200kg * a
∑Fy may = N - 1179 N * cos(56,31) = 0

As for d) The acceleration is constant as the two masses go down they are increasing the speed till a certain time.

e) No clue what formula to use

 

[kuruman]

Thank you for the reply !

I was thinking about this for c)

∑Fx max = 1179 N * sin(56.31) = 200kg * a
∑Fy may = N - 1179 N * cos(56,31) = 0

As for d) The acceleration is constant as the two masses go down they are increasing the speed till a certain time.

e) No clue what formula to use

Please use symbols and substitute the numbers at the very end. It makes reading the equations much easier. I don't know where the 1179 N comes from or what N - 1179 N * cos(56,31) = 0 is meant to express. It doesn't look right. Also, it looks like you call the horizontal direction y and the vertical direction x. That is unconventional and will cause confusion to anyone who might start reading the posts without knowing what you're up to. So please swap the labels so that x is horizontal and y vertical.

Your explanation that the acceleration is constant does not make sense. How can the acceleration be constant, if it is non-zero when the motion starts in part (b) yet in part (e) you are asked to find where the acceleration is zero. Also note that the angle the rope makes with respect to the horizontal decreases as the mass in the middle rises. Furthermore, the tension in the each rope is not equal to the weight of the hanging mass because the rope is accelerating. You need to draw two free body diagrams, one for the rising mass and one for the either dropping mass and get the acceleration as a function of the angle that the ropes make relative to the horizontal (or vertical it doesn't matter which). Then you can study your expression and deduce at what angle the acceleration is zero. Give the matter some serious thought; this problem is not trivial.

 

[Mike94]

Sorry.

Thank you for the explication ! it makes more sense now.

 

[Mike94]

Hello,

I can't still figure it, I guess I'm missing something.

For the rising mass I did:

∑F vertical = mass * acceleration
200 Kg * 9.81 m/s * sin(56.31) = 200 Kg * acceleration
Acceleration = 8.16 m/sec

I tried to do this aswell:

Dropping mass - Rising Mass * sin(tetha)
Dropping mass + Rissing Mass

(400 Kg * 9.81 m/s) - ( 200 Kg * 9.81 m/s ) * sin(56.31)
400 Kg + 200 Kg

Getting an acceleration of: 4.90 m / secDo i have to do * 2 since the rising mass is pulled on two sides ?

∑F vertical = mass * acceleration
( 200 Kg * 9.81 m/s * sin(56.31) ) * 2 = 200 Kg * acceleration
Acceleration = 16.32 m/sec

Thanks again !

 

[haruspex]

200 Kg * 9.81 m/s * sin(56.31)

Please explain how you get that for the net vertical force.

= 200 Kg * acceleration

What else is accelerating?

 

[kuruman]

I can't still figure it, I guess I'm missing something.

You are missing a whole lot because you did not follow my suggestions.
1. You did not use symbolic representation of the variables and parameters. The reason I suggested it is that without it you will not be able to see what's going on.
2. You did not use separate free body diagrams for the cargo mass and for the counterweights. Yes, you have to use two tensions acting on the cargo mass because there are two ropes pulling on it but you cannot assume that the tension is equal to the dropping weight. That's why you need two free body diagrams, to eliminate the tension.
3. You did not take friction into consideration.
4. You assumed that the angle the rope makes with respect to the horizontal is constant and equal to 56.31^o. That's simply not true. As the middle mass rises that angle decreases. That is why you need a symbolic representation to tie it all together.

That's just for starters. What do you think the warning

Warning, the acceleration of the rope is not the vertical acceleration of the cargo!

means? Suppose you draw a pencil mark on the rope and another mark on the cargo mass. Will the accelerations of the two marks be equal? If yes, why? If no, how are they related?

 

[Mike94]

Please explain how you get that for the net vertical force.

I took the weight of the rising mass times the acceleration and the angle of the rope.

What else is accelerating?

The two dropping mass.

 

[Mike94]

You are missing a whole lot because you did not follow my suggestions.
1. You did not use symbolic representation of the variables and parameters. The reason I suggested it is that without it you will not be able to see what's going on.
2. You did not use separate free body diagrams for the cargo mass and for the counterweights. Yes, you have to use two tensions acting on the cargo mass because there are two ropes pulling on it but you cannot assume that the tension is equal to the dropping weight. That's why you need two free body diagrams, to eliminate the tension.

I forgot to upload it. This is where I'm stuck. I though that the Tension is the weight of the dropping mass times the acceleration ( 9.81 m/s ) since there's no angle to the vertical axe.

3. You did not take friction into consideration.

I have to add it to the weight of the rising mass ?

4. You assumed that the angle the rope makes with respect to the horizontal is constant and equal to 56.31^o. That's simply not true. As the middle mass rises that angle decreases. That is why you need a symbolic representation to tie it all together.

I understood that as the rising mass goes up, the angle goes till a certain degree where it stops; under 90 degrees.

That's just for starters. What do you think the warning
means? Suppose you draw a pencil mark on the rope and another mark on the cargo mass. Will the accelerations of the two marks be equal? If yes, why? If no, how are they related?

No, they are related by the force that is pulling down the rope with the cargo mass ?

 

[kuruman]

Are you familiar with the Atwood machine? If not study it here
http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.htmlYour problem is more complicated in that it has two pulleys, the acceleration of the rising mass is not in the same direction as the tension and there is friction. However, this example shows you how to write the equations for the falling masses from their FBDs. Start from there and see if can adapt the example to this problem.

I have to add it to the weight of the rising mass ?

Friction is a force. It belongs in a free body diagram and cannot just be added to a mass.

 

[haruspex]

I took the weight of the rising mass times the acceleration and the angle of the rope.

It helps to get the terminology right. You took the mass of the suspended object and multiplied it by gravitational acceleration (not the actual acceleration) to get the weight of the object. Weight is a force, and you would not be multiplying that by an acceleration.
That much is fine. But how do you justify multiplying that by the sine of that angle? In what direction would such a component of the weight be acting?

I would think you are trying to do ΣF=ma in the vertical direction. The weight of the object is a vertical force, so no need to take a component. What other forces act on the object that have vertical components?