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Two straight ladders, AC and BC, have equal weights $W$. They stand in equilibrium with their feet, A and B, on rough horizontal ground and their tops freely hinged at C. Angle CAB = 60$ ^o$, angle CBA = 30$^o$ and AB = $l$ (see diagram). Find the vertical reactions at A and B.

illustration of mechanical problem

Resolving moments about A:

$$\frac12l\sin30^o\cdot\sin30^o\cdot W + (l-\frac12l\sin60^o\cdot\sin60^o)W = lN$$

$$\implies N = \frac{3W}4$$

Resolving moments about B:

$$\frac12l\sin60^o\cdot\sin60^o\cdot W + (l-\frac12l\sin30^o\cdot\sin30^o)W = lR$$

$$\implies R = \frac{5W}4$$

Find the magnitude and direction of the force exerted on BC by AC.

I couldn't work out how to solve this?

The required answer is $\frac12W$, 30$^o$ above horizontal.

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1 Answer 1

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You'll need to include a vertical and horizontal force at point C due to ladder AC (since we shouldn't assume a direction for the total force -- even though our intuition may prove to be correct). The vertical forces on ladder BC are then related by $F_v + N - W = 0$ and you have already established that $N = \frac{3W}{4}$.

There are now five forces acting on ladder BC, but if we look at the force moments about B, we can neglect the torque due to N and the static friction acting on the foot of the ladder. We can then sum the torques due to W , $F_v$, and $F_h$ to zero. You should indeed find that the total force F at C (using the Pythagorean Theorem) comes to $\frac{W}{2}$ and the direction is 30º above the horizontal (I find it pointing to the right).

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Thanks. I was trying to add a Normal force at C instead and so I couldn't get the right equation, but now it is clear. –  user23060 Apr 12 '13 at 4:22
I had tried to use only a horizontal force at C, as one does with a typical stepladder problem, but the asymmetry in this arrangement makes that approach inadequate. –  RecklessReckoner Apr 12 '13 at 16:14
BTW I got the required answer, but strangely the angle I got was 30deg below the horizontal rather than above. I actually used AC to solve the question (i.e. $R + F_v = W$) because you used BC. –  user23060 Apr 12 '13 at 18:25
I believe that is as it should be: on ladder BC, we have $F_v = \frac{W}{4}$ upward and the torque equation about B leads us to $F_h = \frac{W\sqrt{3}}{4}$ to the right (so we get the total force on BC from AC as $F = \frac{W}{2}$ upward and to the right at a 30º angle.) When we do the same sort of analysis on AC, we have the same magnitudes for $F_v$ and $F_h$ , but these point down and to the left, respectively. So the sum of the two reaction forces of the ladders on each other cancels, as we would expect. –  RecklessReckoner Apr 12 '13 at 18:46
thanks for making that clear –  user23060 Apr 14 '13 at 14:34
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