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Pulley-block system

The figure shows a block (assumed to be a point) being pulled by an ideal string across an elevated pulley. At any time $t$, let the horizontal distance of the block be $x$ from the pulley. The length of the string is $l$ and the height of the pulley is $h$ from the ground. The string is pulled with a speed $u$ as shown. Using calculus I found the speed of the block to be $u \sec \theta$ (where $\theta$ is the inclination of the string with the horizontal) which is given as the correct answer.

So my doubt is: why can't we simply resolve $u$ along the horizontal and say the block's speed is $u \cos \theta$? It doesn't seem wrong to me but isn't correct for some reason. Could someone please explain why?

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  • $\begingroup$ Please post the figure in the question rather than a link to it. $\endgroup$
    – garyp
    May 29 '17 at 18:10
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    $\begingroup$ It is because as well as $l$ the angle $\theta$ changes with time so what you need to do is set up an equation which does not contain $\theta$, $l^2=x^2 +h^2$, and differentiate it implicitly with respect to time. $\endgroup$
    – Farcher
    May 29 '17 at 19:48
  • $\begingroup$ @Farcher I did that and that's how I got u secθ. My question is why we can't resolve u along the horizontal and equate it with v. $\endgroup$
    – Azulene
    May 30 '17 at 3:42
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image

The dots are the points on the string.

The points on the left part of the string are not moving with a uniform velocity $u$. The lines drawn connecting the points depict the displacements of the points on the string. From that, we can see the velocity of every point is not the same. Points on the lower part cover more horizontal distance compared to the points on the upper part. So, you cannot resolve $u$ along the horizontal and say that the velocity of the block is $u \cos \theta$ because the velocity of the string is not $u$.

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    $\begingroup$ Thanks! Your diagram really helped me understand where I went wrong. $\endgroup$
    – Azulene
    May 30 '17 at 15:51
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$x=l \cos \theta$ but $\dfrac{dx}{dt}\ne \dfrac{dl}{dt} \cos \theta$, which is $u\cos \theta$, because the angle $\theta$ is a function of time.

This means that both terms, $l$ and $\cos \theta$, have to be differentiated which gives
$\dfrac{dx}{dt}= \dfrac{dl}{dt} \cos \theta-l\sin \theta \dfrac{d \theta}{dt } =u \cos \theta-l\sin \theta \dfrac{d \theta}{dt }$

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Because $u$ is only a component of the block's velocity ( along $\hat{r}$) it also has a component along $\hat{\theta}$. The sum of the two along the vertical is zero and along the horizonal gives the speed of the block. (If it was the only component, then the block would fly, hence it obviously has a component along $\hat{\theta}$)

Another way to look at it would be $x = lcos\theta$ but since $\theta$ is a function of time you can't differentiate this equation to get $v$

You could however resolve the block's velocity (which is entirely along x) along and perpendicular to the string and equate the component along the string to $u$ which gives the desired result.

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Cosθ=(length of X/length of L). You need to find (delta length X/delta length L) or dx/dl. One is a relationship between lengths, the other is a relationship between changes in length. The two are just different.

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  • $\begingroup$ If that's so, why do we resolve any variable of linear motion into its components? $\endgroup$
    – Azulene
    May 30 '17 at 3:44

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