3 masses attached by rope on table with 2 pulleys.

In summary: And since the friction force is not balanced by any other force, there must be a net force in the horizontal direction.F = maT2 - Ff = ma - this is the equation for M3T2 = m3a + FfT2 = (2kg)(2.1 m/s^2) + 4.704 NT2 = (4.2 N) + 4.704 N = 8.904 NI think that's the right answer, but I'm not sure.
  • #1
MaximaMan
10
0

Homework Statement



You have 3 masses. M1 = 4 kg. M2 = 1kg. M3 = 2kg, M1 is attached to M2 by a rope. M2 is attached to M3 by a rope. M1 is hanging from a pulley attached to the left end of a table, M2 is on the table, and M3 is hanging from a pulley attaced to the right end of the table.

The coefficient of friction for the table is 0.48

Find the downward acceleration of M1, the leftward acceleration of M2 and the upward acceleration of M3.

Also find the Tensions of the left rope(the one attaching M1 and M2) and the right rope(the one attaching M2 and M3)

KNOWN:
  • Acceleration for M1, M2 and M3 are the same, right?
  • The masses are moving, therefore the net force /= 0, right?

Homework Equations



Ff = mu k * FN

F = ma

The Attempt at a Solution



I don't understand this at all.

Ff = (0.48) (9.8) (1kg) = 4.704

FnetM1 = 19.6 - 4.704 = 15.096 N (to the left).

Acceleration = 15.096 / mass(1kg) = 15.1 m/2

This is wrong. I can't figure anything out...for Gods sake someone please help me.
 
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  • #2
FnetM1 = 19.6 - 4.704 = 15.096 N (to the left).
This looks correct, now just find the acceleration.
It would look a little nicer this way:

The sum of the forces = Ma. Goes left so that's positive. Friction force opposes.
m1g - 4.704 - m3*g = Ma
 
  • #3
Ok calm down MaximaMan :D. Ok so firstly have you draw a force diagram with all the relevant forces and any resultant accelerations? it is vitally important that you get a good diagram draw as it really helps to visualize the problem.

Now in order to solve this problem you will end up with a system of equations, it actually is as simple as writing out the equations to describe the forces and any resultant forces on all of the masses. You have to also realize something important about the acceleration on all the masses, and this line "Find the downward acceleration of M1, the leftward acceleration of M2 and the upward acceleration of M3" is almost there to throw you off.

What you currently missing in you equations it the tensions (which is why a diagram will help) draw what is depicted from the question, draw the weights acting on the masses, and also the relevant tensions in the strings.
 
  • #4
Yeah I got that. But which mass do I use? The sum of M1,2,3? Or M1?

Ive tried a couple of different answers and none of them were right(though I don't have the right one).

Galadirith, the problem actually asks for what direction each mass is moving. I just filled it in here for simplicity sake(m1 down, m2 left etc etc). I have drawn this problem out, like 4 different ways. Each mass as its own free body diagram, and also as the whole diagram. I still can't sort it out to get this...
 
  • #5
The M in Ma is the total of all the moving masses.
Are you getting around 2 m/s^2?
 
  • #6
I put in 1.9 and it told me my answer was not within 10% of the correct answer.
 
  • #7
Wait. Okay, so somehow, I managed to get 1.9 something, but you're correct the answer is 2.1xxxx. I put it in and got it right.Now onto the tension. For the left rope, isn't it simply T1 = m1*g and T2 = m2*g?Galadirith, my professor said for the sake of life, that we are always to consider the string to be massless and rigid(aka, it can't stretch).
 
  • #8
Oh MaximaMan of course, I didnt mean to imply anything about the properties of the ropes themselves. With you equations of the tensions in the strings, remember that M1 and M3 are also accelerating.
 
  • #9
T1 is the force needed to overcome the friction plus the gravity on M3 plus the ma needed to accelerate m2 and m3.

Galadarith, I would be most interested in seeing the solution with the tensions in there from the start. It seems unnecessary to me, but there would be a certain elegance to having all the forces in the first line of the solution. Would you have an equation for each mass - a system of 3 equations to solve to find the acceleration and the two tensions?
 
  • #10
Delphi51, yes basically, and I want to retract my previous comment, you were absolutely right I totally confused what was going on.
 
  • #11
Delphi51 said:
T1 is the force needed to overcome the friction plus the gravity on M3 plus the ma needed to accelerate m2 and m3.

T1 = 4.704 Ff + 2(9.8) + (1kg)(2kg)(2.1 m/s2) = 30.6 N right? - correct answer, thanks guys!

However! I calculated T2 the same way, and didnt get the right answer. I got 54.4 N and that is incorrect. Any ideas?
 
Last edited:
  • #12
T2 is just the force of the rope on m3, causing it to accelerate at 2.13. Just use F=ma.

Hey, Galadarith, thanks for the tutoring on tension forces! I think I understand how to incorporate them at the beginning now. I'm a retired high school teacher trying to figure these questions out just for brain exercise, and of course I really enjoy helping kids.
 
  • #13
Well I ran out of chances to get the right answer for T2.

I did f=ma and the answer was wrong. I swear...lol. Anyhow, I did it F = (2kg) (2.1 m/s2) and I tried F = (2kg) (9.8 - 2.1) thinking that M3 had the acceleration of gravity and the upward acceleration. (I don't know why I thought this, just trying to use my head! Though, half the time I over think things, thinks IB!)

I am still interested in the answer for learning purposes though.
 
  • #14
Since there is an acceleration, the net force cannot be zero. Don't forget that the only time an object can be in motion with a net force of zero is if there is no acceleration.
 

Related to 3 masses attached by rope on table with 2 pulleys.

What is the purpose of the 3 masses attached by rope on a table with 2 pulleys?

The purpose of this setup is to demonstrate the principles of pulleys and how they can be used to multiply and redirect forces in a system. It also allows for the study of the effects of different masses and rope lengths on the overall system.

How do the pulleys affect the overall force in the system?

The pulleys act as a mechanical advantage, reducing the amount of force needed to lift the masses. As the rope is redirected through the pulleys, the force required to lift the masses is divided between the multiple ropes, making it easier to lift the masses.

What variables can affect the behavior of the system?

The behavior of the system can be affected by variables such as the masses of the objects, the length and tension of the ropes, the friction of the pulleys, and the angle at which the ropes are attached to the pulleys. Changing these variables can alter the overall force and movement of the system.

How does the number of pulleys affect the system?

The number of pulleys affects the mechanical advantage of the system. As more pulleys are added, the force required to lift the masses is further reduced. However, adding more pulleys also increases the complexity and potential for friction in the system.

What real-world applications can this system be used for?

This system can be used in various real-world applications, such as elevators, cranes, and sailboats. It allows for the efficient transfer of force and can be used to lift heavy objects with less effort. Understanding the principles behind this system can also aid in the design and improvement of mechanical systems.

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