The question:


In a different arrangement, the string is cut so that the lengths of the two parts are $0.5m$ and $2.3m$. Describe how the block hangs in equilibrium in this case and state the tensions in the two strings.

My attempt:

hand drawn version

I used $a^2 = b^2 + c^2 - 2bc\cos{A}$ with $A =\alpha$: $$2.3^2 = 2^2 + 0.5^2 - 2(2)(0.5)\cos{\alpha}$$ $$\cos{\alpha} = \frac{2.3^2-(2^2+0.5^2)}{-2(2)(0.5)}= \frac{1.04}{-2} = -0.52$$ $$\alpha = \arccos{(-0.52)} = 121.33^\circ$$

However the answer to this question states:

Tension in AC is 50 N (it takes all the weight)

Tension in BC is zero (it is slack)

Which implies $\alpha = 90^\circ$ as Tension in $AC$ only has a vertical component

What am i doing wrong?


Since the cosine of $\alpha$ is negative, $\alpha$ is an obtuse angle, so the diagram's a bit of a con: to have both strings taught, $AC$ will be beyond the vertical, further to the left. But that doesn't make physical sense: $BC$ can't provide a force to hold the block beyond the vertical line through $A$ because it's a string. Therefore BC is actually slack, and $AC$ is vertical.

  • 3
    $\begingroup$ Basically, this is not just a question of being able to do the calculations, but also of being able to interpret the results and evaluate whether a calculated solution makes sense. Certainly, if they had been rods and not ropes, you would get the configuration with $\alpha = 121.33^\circ$. $\endgroup$ – Arthur Jun 15 '17 at 13:13
  • $\begingroup$ thank you, when you wrote $AB$ did you mean $AC$ ? $\endgroup$ – Sonny Da Silva-Peters Jun 15 '17 at 13:14
  • $\begingroup$ @SonnyDaSilva-Peters I meant $BC$, in fact. $\endgroup$ – Chappers Jun 15 '17 at 13:16
  • $\begingroup$ @Chappers but surely $BC$ could provide a force beyond the vertical line $A$ isn't it $AC$ that cant provide a force if its angle i obtuse? $\endgroup$ – Sonny Da Silva-Peters Jun 15 '17 at 13:18
  • $\begingroup$ If you look at the forces on the block with both strings going to the right (as they will if $\alpha$ is obtuse), because both the forces are tensions, they both point to the right and you can't have equilibrium. $\endgroup$ – Chappers Jun 15 '17 at 13:21

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