Consider the the following problem: enter image description here

Here is a force diagram showing the situation: enter image description here

For part (i) Taking moments about $B$ for $BC$ gives

$84.5Lcos\beta=2LT$ So $T=39$N

For part (ii) Resolving vertically upwards on $BC$ gives

$39\cos\beta -84.5=-r$ which results in $r=48.5$N in the direction shown on diagram (downwards). I'm aware that the question asks for the forces acting on $BC$ so I deduced that by newtons third law $Y=48.5$N (upwards). Everything so far is correct but here's the problem:

To find the horizontal component I resolved leftward on $BC$ to obtain $p=39\sin\beta$ giving $p=15$N direction as shown in diagram (leftward). Again realizing that the question asks for forces on $BC$, by newtons third law $X=15$N (rightward): Here's the official answer:

enter image description here

The marks-scheme is always correct. So could someone please kindly explain what I did wrong?


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  • $\begingroup$ In statics, the point at which moments are taken does not matter, but if you switch the sense of the $x$ axis for example, you have to switch the sense of what a positive moment is also. $\endgroup$ – ja72 Sep 3 '14 at 23:22
  • $\begingroup$ @ja72 whichever you like? makes no difference. $\endgroup$ – BLAZE Sep 3 '14 at 23:59

For (ii) force $p$ acts to the left. This is obvious from statics since $T$ acts to the right in order to have balance.

The diagrams has the assumed positive directions for $p$ and $r$ and it is confusing that the answer talks about $X$ and $Y$.

Using $\cos \beta = \frac{12}{13}$, $\sin \beta = \frac{5}{13} $ and $W=84.5\,{\rm N}$

$$ \left. \begin{aligned} 2 L T & = L W \cos \beta \\ T \sin \beta & = p \\ T \cos \beta - W &= r \end{aligned} \right\} \begin{aligned} T &= \frac{W}{2} \cos\beta \\ p & =\frac{W}{2} \cos\beta \sin\beta \\ r & = \frac{W}{2} \cos\beta \cos\beta - W \end{aligned} $$

which makes $p=15$ positive (and hence to the left) and $r=-48.5$ negative (and hence upwards).

Again from inspection you could have guessed the directions since the forces must oppose the tension $T$.

  • $\begingroup$ Yes the X and Y are just the marks-schemes way of putting the horizontal and vertical components. Do you agree with the their answer or mine? $\endgroup$ – BLAZE Sep 3 '14 at 23:38
  • 1
    $\begingroup$ I am looking at a more detailed answer now. I am trying to deduce if I have to use $\sin \beta = \sqrt{1-\cos^2\beta}$ or would in cancel out. $\endgroup$ – ja72 Sep 4 '14 at 0:01
  • $\begingroup$ For clarity I have changed the q and t to X and Y $\endgroup$ – BLAZE Sep 4 '14 at 0:06
  • $\begingroup$ Okay thanks for the verification, so the mark-scheme answer really is wrong? $\endgroup$ – BLAZE Sep 4 '14 at 0:44
  • $\begingroup$ The question asks for forces acting on BC at B and hence looking for $p$ and $r$, not $X$ and $Y$. So the answer is correct. $\endgroup$ – ja72 Sep 4 '14 at 0:50

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