What is the speed of M2 at t=1.75 s in a system of three blocks and two pulleys?

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Homework Help Overview

The problem involves a system of three blocks and two pulleys, where the blocks are released from rest, and the goal is to determine the speed of the middle block (M2) at a specific time after release. The setup assumes no friction and massless, frictionless pulleys, with given masses for each block.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations of motion for each mass, focusing on the tension in the strings and the acceleration of the system. There are attempts to derive the acceleration and velocity of M2 based on the forces acting on the blocks.

Discussion Status

Several participants have shared their equations and attempted to solve for acceleration and tension. Some have expressed confusion regarding the differences in tension for the two strings and how to correctly apply the equations. There is a mix of successful and unsuccessful attempts to find the correct values, with some participants confirming their results while others continue to seek clarification.

Contextual Notes

Participants note discrepancies in their calculations and results, with some indicating that they arrive at zero for acceleration under different mass configurations. The discussion reflects ongoing exploration of the problem setup and assumptions regarding the system's dynamics.

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Homework Statement


three blocks 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 = 3.0 kg, M2 = 9.0 kg, M3 = 5.0 kg.

M2 is the middle block connected to M1 (to the left) and M3 (to the right)

Calculate the speed of M2 at a time 1.75 s after the system is released from rest.


Homework Equations


Since they will all have the same acceleration, i was trying to look for a and then multiply that by the t.



The Attempt at a Solution



I set up this equation:
FM1 = Tension - M1*g = M1a
FM3 = Tension - M3*g = (M1 + M2 + M3)a

and i get a to be 1.4m/s^2 and Vf to be 2.45m/s but it is incorrect. Any help pls?
 
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Tension in the two strings are not the same.
Note down which mass goes up and which mass goes sown.
Accordingly write down three equations for three masses.
You have three unknowns, T1, T2 and a. Solve for them.
 
wait, how are the two tensions different though?
I know that FM1 = Tension - M1*g = M1a is correct, so is tension for Fm3 going to be (m1+m2)a?
 
M3*g - T1 = M3*a ...(1)
T1 - T2 = M2*a...(2)
T1 - M1*g = M1*a ...(3)
Solve for a.
 
if i found a, how do i find the speed of m2?
 
leavex said:
if i found a, how do i find the speed of m2?

Use v = vo + at. vo is zero.
 
i tried it doesn't work, i have the same question but but different m
m1=2 m2=8 m3=4
i get zero for answer and its wrong... i did exactly the same way you did above...
 
leavex said:
i tried it doesn't work, i have the same question but but different m
m1=2 m2=8 m3=4
i get zero for answer and its wrong... i did exactly the same way you did above...

What is your value of a?
 
i solved id and got 0 for a
m3g-m1g-m1g=4-2-2=0
 
  • #10
4g - T2 = 4a...(1)
T2 - T1 = 8a...(2)
T1 - 2g = 2a...(3)

From these equations find a.
 
  • #11
i did it doesn't wark i just don't know what do i do wrong...
i got a=1.962
t= 1.45
v=2.84 m/s
why is it wrong??
 
  • #12
If you add (1) and (3), you get

T1 - T2 + 2g = 6a.

2g = 6a + T2 - T1 = 6a + 8a = 14a.

So a = 2g/14 = 2 * 9.8/14 = 1.4 m/s^2

Now find the velocity of m2.
 
  • #13
yeahhhhhhhhhhh it fianely worked, thank you so mach!
 

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