# Calculating distance travelled in two-pulley system as a consequence of surface object moving

Using this two-pulley system, if the block moves 1 cm to the right, how far will the grapes move down?

So I'm not sure if I'm screwing up the first part here, because I'm not getting a 'numerical' answer when I believe I should be.

To transcribe my mathematical work:

$$m_1 = 4, m_2 = 0.8$$ $$F_{x_1}=2T=m_1a_{1_x}\implies a_{1_x}=\frac{2T}{m_1}=\frac{2T}{4}=\frac{T}{2}$$

$$F_{y_2}=m_2g-T=m_2a_{2_y}\implies a_{2_y}=\frac{m_2g-T}{m_2}=g-\frac{T}{m_2}=10-\frac{5T}{4}$$

$$x_1=x_{1_0}+v_{1_{x_0}}+\frac{1}{2}a_{1_x}t^2 \implies x_1 =\frac{1}{2}a_{1_x}t^2\implies \frac{1}{100}=\frac{1}{2}\left( \frac{T}{2} \right)t^2 \implies t^2=\frac{1}{25T}$$

$$y_2=y_{2_0}+v_{2_{y_0}}+\frac{1}{2}a_{2_y}t^2 \implies y_2 = \frac{1}{2}a_{2_y}t^2 \implies \frac{1}{2} \left(10 - \frac{5T}{4} \right)\left(\frac{1}{25T} \right) = \frac{1}{2} \left(\frac{10}{25T} - \frac{5T}{100T} \right)=\frac{T}{5}-\frac{1}{20}$$

The bunch of grapes travels $$20T-5$$ centimeters if the block travels $$1$$ centimeter. What?

You are hiring an excavator to bury a hamster.

It's a kinematic problem, not a dynamic problem. It doesn't matter what are the masses or how large is the gravity. All you need is the kinematic constraint that the length of a rope is constant. The length of a rope is a sum of three segments: from gray hook to blue pulley (1), from blue pulley to purple pulley (2) and from purple pulley to grapes (3). When the block moves right 1 cm, both $$l_1$$ and $$l_2$$ decrease by 1 cm. And since $$l_1+l_2+l_3=\mathrm{const}$$, we conclude that $$l_3$$ should increase by 2 cm.

You cannot solve this problem dynamically without one more equation to determine $$T$$. Any massless rope will have the uniform tension, so all of your equations will be true. However, if you imagine the opposite, an absolutely elastic massless rope, it won't resist stretching, so $$T=0$$ and the grapes will be falling endlessly without block moving. In inelastic case, “one more equation” is the kinematic constraint. You conclude that since $$l_1+l_2+l_3=\mathrm{const}$$, $$2a_{x1}=a_{y2}$$, this allows you to find $$T$$ and get the numerical value. However, it's kind of doing a lot of unnecessary work for free.

• oh my God, @Vasily Mitch, thanks - I completely overlooked that the string's length stays constant. However, with my calculations, could I theoretically use that to calculate the tension $T$ in the string as well? $\frac{T}{5} - \frac{1}{20} = 2$ means $T = \frac{41}{8}$ when the block moves $1$ cm. Is this mathematically valid too? Oct 18, 2021 at 22:04
• I have added the seconf paragraph Oct 18, 2021 at 22:08
• And in addition, you have an algebraic mistake, it should be $\frac1{5T}-\frac1{40}$ Oct 18, 2021 at 22:17
• @chuloobydooby The tension in the rope has nothing to do with the fact that the block moves 1 cm. You cannot use the 1 cm dispalcement to find the tension. The first part of the question was just a step meant for you to realize the relationship between accelerations explained by Vasily Mitch.
– nasu
Oct 19, 2021 at 4:02