A pulley is pulled with external force $F$. $x$ and $y$ denote the displacement of two ends of strings of the pulley and $z$ is the displacement of the pulley.

Prove That $$z = \dfrac{x+y}{2}$$

enter image description here

My teacher stated it in class without providing a proof. I think that it is valid for ideal conditions (massless pulley and string) and only with certain kinds of forces, but he says it is valid in all cases.

Can anyone help me with the general proof? Thanks in advance.

  • 1
    $\begingroup$ @AnubhavGoel Care to elaborate? $\endgroup$
    – Henry
    Commented Mar 21, 2016 at 17:01
  • $\begingroup$ The only condition you need for the validity of the relation is that the string is inextensible. I suggest you to start looking at particular cases. For example, if z=0 it should be x=-y. And if x=y then z=x=y. $\endgroup$
    – GCLL
    Commented Mar 21, 2016 at 22:32
  • 1
    $\begingroup$ Hi Henry and welcome to the Physics SE! Please note that this is not a homework help site. Please see this Meta post on asking homework questions and this Meta post for "check my work" problems. $\endgroup$ Commented May 29, 2016 at 7:41

2 Answers 2


enter image description here

Let l is length of string. m is length of pulley string. h is height of system. k is height of pulley after it is raised.

$h = m+ \frac{l}{2} \tag{1} $

$k = \frac{l+x+y}{2} \tag{2} $

$ h+ z = m +k $

From (2) $h+z = m+ \frac{l+x+y}{2} $

From (1) $ m+ \frac{l}{2} +z = m+ \frac{l+x+y}{2} $

On Solving



If the length of the string is fixed (i.e. if the string is inextensible), then we have:

($l$ is the length of the string) enter image description here

$$l=(s_1-s_3)+(s_1-s_5)+\pi R=2s_1-s_3-s_5+\pi R$$ $$l=(s_2-s_4)+(s_2-s_6)+\pi R=2s_2-s_4-s_6+\pi R$$ $$z=s_2-s_1\quad x=s_4-s_3\quad y=s_6-s_5$$ $$\Longrightarrow\; 2s_2-2s_1=(s_4-s_3)+(s_6-s_5)$$ $$\Longrightarrow\; 2z=x+y$$


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