Determine the speed of the object $M$ as a function of $x$, if the right part of the line ($B$) moves horizontally with a constant velocity $v_{0}$ and if $B$ and $M$ coincide when $x = 0$.

This is from a first year single variable calculus textbook that does not treat physics or engineering.

Attempted solution:

Here is an image of the problem:

enter image description here

From the question, we know that the length of the string is $2h$. The amount that goes from B to the top (let us call that c) can be found with the Pythagorean theorem:

$$c^2 = h^2 + x^2 \rightarrow h = \sqrt{c^2-x^2}$$

Since we know that:

$$c+h = 2h$$

We have:

$$h = \sqrt{h^2-x^2}$$

Taking the derivative:

$$\frac{dh}{dt} = \frac{1}{2 \sqrt{h^2-x^2}} 2h \frac{dh}{dt}$$

However, this does not take $v_{0}$ into account. Since it is in the direction to the right instead of up (which the mass experiences), there has to be some angle to work with:

$$\tan \theta = \frac{h}{x}$$

However, I am a bit stuck here. Do not seem to find any productive way to introduce $v_{0}$.

  • 1
    $\begingroup$ In your equations you are using $h$ as the length of the string from the pulley to $M$. This is not correct as it is defined as the fixed length from the pulley to the bottom line. In particular, your second line says $c=h$, which is only true when $x=0$ $\endgroup$ Aug 1, 2016 at 15:02

1 Answer 1


Let $y$ be the distance from the pulley to $M$. As you say, $c=\sqrt {x^2+h^2}$ and the length of the string is $2h$, so $y=2h-c=2h-\sqrt {x^2+h^2}$. Now you can take $\frac d{dt}$ of both sides, then set $\frac {dx}{dt}=v_0$


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