3 Sliding Masses; find speed

  • #1

Homework Statement


The three blocks shown are relased at t=0 from the position shown in the figure. Assume that there is no friction between the table and M2, and that the two pulleys are massless and frictionless. The masses are: M1 = 1.0 kg, M2 = 7.0 kg, M3 = 3.0 kg.
Calculate the speed of M2 at a time 1.45 s after the system is released from rest.
(M1 is on left side of box, M2 is on the top of the box, M3 is on the right side of the box. Corners of box are pulleys.)


Homework Equations


None.



The Attempt at a Solution


I have ZERO clue.
 

Answers and Replies

  • #2

Homework Statement


The three blocks shown are relased at t=0 from the position shown in the figure. Assume that there is no friction between the table and M2, and that the two pulleys are massless and frictionless. The masses are: M1 = 1.0 kg, M2 = 7.0 kg, M3 = 3.0 kg.
Calculate the speed of M2 at a time 1.45 s after the system is released from rest.
(M1 is on left side of box, M2 is on the top of the box, M3 is on the right side of the box. Corners of box are pulleys.)


Homework Equations


None.



The Attempt at a Solution


I have ZERO clue.
These problems are always very interesting ones.

First, you have to draw a FBD for all three masses.
 
  • #3
3,816
92
Where is the figure? :confused:
 
  • #6
So basically, there are two pulleys attached to the box.

There is a single, massless string run through the pulley system.

M1 is hanging on the left side of the box and M3 is hanging on the right side of the box (attached to string I presume).

M2 has a string connected on both sides? Yes?
 
  • #7
The image should be 3boxes on a pulley system.
 
  • #8
Yes, Aggression, that is right.
 
  • #9
3,816
92
The image should be 3boxes on a pulley system.

Is that a university website or something like that?
If so, i think i can't access that until i am a student of it.
 
  • #10
Alright, now we can get this problem started.

So, you'll have 3 separate FBDs.

M1's FBD will have a T force going up and a Weight Force going down.

M2's FBD will have tension forces on both sides, a normal force going up that is equal to a weight force going down.

M3's FBD will have a T force going up and a weight force going down.

Someone correct me if I'm wrong.
 
  • #11
Image should be attached.
 

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  • #12
PeterO
Homework Helper
2,433
58

Homework Statement


The three blocks shown are relased at t=0 from the position shown in the figure. Assume that there is no friction between the table and M2, and that the two pulleys are massless and frictionless. The masses are: M1 = 1.0 kg, M2 = 7.0 kg, M3 = 3.0 kg.
Calculate the speed of M2 at a time 1.45 s after the system is released from rest.
(M1 is on left side of box, M2 is on the top of the box, M3 is on the right side of the box. Corners of box are pulleys.)


Homework Equations


None.



The Attempt at a Solution


I have ZERO clue.

consider the 3 masses plus string as a single sytem, with the pulleys just adjusting a couple of directions.

The Force trying to move the system clockwise is the weight of the right hand mass.
The force trying to move the system in the anti-clockwise direction is the weight of the left hand mass.
One of those is bigger and "wins".
The net force [one weight minus the other] acts on the whole system. F = ma is then used.
 
  • #13
But i don't have the acceleration of the force...
 
  • #14
3,816
92
consider the 3 masses plus string as a single sytem, with the pulleys just adjusting a couple of directions.

The Force trying to move the system clockwise is the weight of the right hand mass.
The force trying to move the system in the anti-clockwise direction is the weight of the left hand mass.
One of those is bigger and "wins".
The net force [one weight minus the other] acts on the whole system. F = ma is then used.

PeterO, please don't directly come on F=ma relation. It would be better if you help the OP make equations for each body. :smile:
 
  • #15
Okay, can you give me a hint please?
I'm sorry, but i'm really bad at physics and I'm stressed and Idk where to go.
 
  • #16
Well, after you draw your FBDs, you will have a system of 3 equations.

So, sum the forces in each situation.

We will denote T[itex]_{1,2}[/itex] as the tension force between the M1 and M2 object.
We will denote T[itex]_{2,3}[/itex] as the tension force between the M2 and M3 object.
T[itex]_{2,1}[/itex] = T[itex]_{1,2}[/itex]
T[itex]_{3,2}[/itex] = T[itex]_{2,3}[/itex] By Newton's 3rd law.


Now, sum of the forces for M1....

[itex]\Sigma[/itex]F = T[itex]_{1,2}[/itex] - m[itex]_{1}[/itex]g = m[itex]_{1}[/itex]a

Sum of the forces for M2.....

[itex]\Sigma[/itex]F = - T[itex]_{2,1}[/itex] + m[itex]_{2}[/itex]g - m[itex]_{2}[/itex]g + T[itex]_{2,3}[/itex] = m[itex]_{2}[/itex]a

Sum of the forces for M3....

[itex]\Sigma[/itex]F = -T[itex]_{3,2}[/itex]+m[itex]_{3}[/itex]g = m[itex]_{3}[/itex]a


I'm not 100% sure that these are correct. Someone double check me.
 
  • #17
MEH! I have no clue!
 
  • #18
Let's assume they are. haha

Now, add up all the equations. The tension forces will cancel leaving you with.....

[itex]\Sigma[/itex]F = m[itex]_{3}[/itex]g - m[itex]_{1}[/itex]g = m[itex]_{1}[/itex]a + m[itex]_{2}[/itex]a + m[itex]_{3}[/itex]a

So, factor out a g from the left side, and an a from the right side.....

g(m[itex]_{3}[/itex]-m[itex]_{1}[/itex]) = a(m[itex]_{1}[/itex] + m[itex]_{2}[/itex] + m[itex]_{3}[/itex])

Solve for a and that will give you the acceleration of each block.

m1 = 1.0 kg
m2 = 7.0 kg
m3 = 3.0 kg
 
  • #20
m2g canceled out because the equation for M2 included.....

m2g - m2g = 0

So, that's why they canceled out.
 
  • #21
It doesn't matter, i timesed that by the time and i got the answer :d
 

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