Solving Acceleration of 4kg Mass in Pulley System

Bucky
Messages
79
Reaction score
0

Homework Statement


One end of a light inextensible string is attatched to a 4kg mass on a smooth horizontal table. The string passes over the edge of the table and to it's other end is attatched a light smooth pulley. Over this pulley passes another light inextensible string, with masses of 3kg and 2kg attatched to it's ends. If the system is allowed to move freely, show that the 4kg mass moves with acceleration of 6g/11.


Homework Equations





The Attempt at a Solution



http://img247.imageshack.us/img247/8569/diagramjy9.th.jpg

I'm having a bit of a problem from the outset getting the equations of motion. I don't think the diagram is correct. I feel like there should be another acceleration representing the movement of the pulley against the 4m mass, but I'm not sure how to represent it (and part of me thinks that it's covered by the tension t1. Any help on setting out the problem would be appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
Are you trying to use a Legrangian, is that what you mean by equations of motion? Yes, the mass on the table will accelerate too.
 
Mindscrape said:
Are you trying to use a Legrangian, is that what you mean by equations of motion? Yes, the mass on the table will accelerate too.


but how does that fit into the equations of motion?
from what i can see the tension 'connects' the 4m mass to the pulley system.

here are my equations of motion...i know the're wrong but i don't see how the acceleration of the 4m mass on the table fits in...the questions up till now have just been systems of pulleys with weights/more pulleys on them, we haven't encountered a mass on a table before.


4m mass on table: 4m ? = 4mg - t1
3mg mass : 3ma = t2 - 3mg
4mg mass : 4ma = 4mg - t2
 
Hello,

There are three forces exerted on the 4mg on the table:
4mg, normal force by the table, t1.
The first two forces are balanced off in the vertical direction.
The net force on the 4mg is only the t1 in the horizontal direction.
Therefore, the first one of your equations shlould be
t_1=4m\times a_1

Do you think the 3m and 4m on the pulley have accelerations of the same magnitude but opposite direction ?


Best Regards
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...

Similar threads

Back
Top