2
$\begingroup$

Atwood machine

my attempt

enter image description here

$$4(x_3-c_1)+x_p=L$$ which give $4*a_3=-a_p$
such that $a_p$ acceleration of pulley
how to get relation between $a_1$ and $a_2$ and $a_3$ using conversion of string and how to solve the same problem by work done by tension Why is the net work done in a pulley-string system zero?

$\endgroup$
4
$\begingroup$

Instead of thinking about accelerations $a_1$, $a_2$, and $a_3$, it's easier to think about displacements $x_1$, $x_2$, and $x_3$. Furthermore, imagine replacing the string with a rubber band, which can freely stretch or shrink, so you can imagine one mass moving at a time.

Given a displacement $x_1$, the rubber band shrinks by $2x_1$. Similarly, given a displacement $x_2$, the rubber band shrinks by $2 x_2$, and given displacement $x_3$ it shrinks by $x_3$. But since we actually have a string, the total length of the string must stay constant, so $$2x_1 + 2x_2 + x_3 = 0.$$ By differentiating both sides with respect to time twice, we have $$2a_1 + 2a_2 + a_3 = 0.$$ This is the desired relation between the accelerations, using conservation of string.

$\endgroup$
  • $\begingroup$ This rubber band mental picture is very helpful! Thank you. $\endgroup$ – Macrophage Feb 15 '20 at 20:33
1
$\begingroup$

(Not an answer)

How did you come up with that equation? If you see how long the string is (from left to right) you have that:

$ L=x_p+(x_p-c_1)+(x_p-c_1)+(x_p-c_1)+(x_3-c_1)=4(x_p-c_1)+x_3 $

Now this makes sense because when $ x_p=c_1 $ you have that $ x_3 = L $.

$\endgroup$
1
$\begingroup$

Try starting with a free-body diagram of each mass individually without any assumptions about the pulleys or acceleration and you will find: $$ \begin{align} \mathbf{T}_{1a} + \mathbf{T}_{1b} - m_{1} g & = m_{1} \ \mathbf{a}_{1} \tag{1a} \\ \mathbf{T}_{2a} + \mathbf{T}_{2b} - m_{2} g & = m_{2} \ \mathbf{a}_{2} \tag{2a} \\ \mathbf{T}_{3} - m_{3} g & = m_{3} \ \mathbf{a}_{3} \tag{3a} \end{align} $$

Now if we assume the pulleys are massless and frictionless, then we know that no energy/momentum is lost to the generation of rotation of the pulleys or dissipation (e.g., heat generated by frictional forces). This results in: $$ \begin{align} \mathbf{T}_{1a} & = \mathbf{T}_{1b} = \mathbf{T}_{1} \tag{1b} \\ \mathbf{T}_{2a} & = \mathbf{T}_{2b} = \mathbf{T}_{2} \tag{2b} \\ \mathbf{T}_{3} & = \mathbf{T}_{2} \tag{3b} \end{align} $$ We can take this one step further and say that all tensions are equal if the rope is incompressible and infinitely strong, i.e., it will not stretch, so that $\mathbf{T}_{1} = \mathbf{T}_{2} = \mathbf{T}_{3} = \mathbf{T}$.

Each free-body diagram is a 1D situation, so we can remove the vector (bold) notation and replace with the sign of each quantity. If we pretend like we do not know the sign of each acceleration and retain unique masses, then we solve for $T$ in each to find: $$ \begin{align} T & = \frac{ m_{1} }{ 2 } \left( g \pm a_{1} \right) \tag{1c} \\ T & = \frac{ m_{2} }{ 2 } \left( g \pm a_{2} \right) \tag{2c} \\ T & = m_{3} \left( g \pm a_{3} \right) \tag{3c} \\ \end{align} $$ This is three equations with four unknowns (i.e., $a_{1}$, $a_{2}$, and $a_{3}$), if we assume that $m_{1} = m_{2} = m_{3} = m$. The last equation is given in @Kirtpole's answer, where from kinematics we can see that the ratio of accelerations must balance the ratio of displacements (i.e., sometimes called the conservation of string method).

The rest is just solving for each acceleration, which is just symbol gymnastics.

$\endgroup$
  • $\begingroup$ I want to solve it by conservation of string method $\endgroup$ – Abdelrhman Fawzy Dec 19 '18 at 13:25
  • 1
    $\begingroup$ You cannot use conservation of string alone, as this will just give relative ratios. Regardless, I added a note explaining how the conservation of string method comes into answering the question. $\endgroup$ – honeste_vivere Dec 19 '18 at 15:12
1
+50
$\begingroup$

enter image description here

OBSERVATION:

  • P2,P4 are attached to the ceiling,so they don't move

NOW,

how to get relation between a1 and a2 and a3 using conversion of string

Conservation of length of string also necessarily implies that total change in the length of the string is Zero,you've got to agree!

Let P1,P3 and the mass having the acceleration $a_3$ move upwards(as you've given in the question that their acceleration directions are upwards)by distances $x_1$,$x_2$,$x_3$ respectively.

Change in length of string AB: -$x_1$(negative because there is a decrease in length)

Change in length of string CD: -$x_1$

Change in length of string FE: -$x_2$

Change in length of string HG : -$x_2$

Change in length of string OI : -$x_3$

TOTAL CHANGE IS ZERO:

(-$x_1$) + (-$x_1$) + (-$x_2$) + (-$x_2$) + (-$x_3$) = 0

$2x_1$+ $2x_2$+$x_3$=0

To get acceleration you've to differentiate twice w.r.t time

NOTE: we've already considered the signs

So,the answer you are looking for: $$2a_1 + 2a_2 + a_3 = 0.$$

$\endgroup$
0
$\begingroup$

If the pulley is real then work done by tension is not zero. If pulley is ideal then work done by tension is zero. Consider a system of a single pulley attached with two masses, Let the displacement of 1st body be x then by conservation of length of string the displacement of 2nd body will be in opposite direction with respect to 1st body but equal in modulus, Let the tension in the string at both ends be T{not a rotational system of pulley}, Work done by tension in displacing 1st body=T.Xcos0°=TX. Work done by tension in displacing 2nd body=T.Xcos180° =--TX So,total work done by tension=TX --TX=0. We can generalise this relation for any system of pulleys i.e. total work done by tension=0{only for ideal pulleys}.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.