A steel girder AB has weight 210 N - Edexcel - A-Level Maths Mechanics - Question 5 - 2006 - Paper 1

Using the equality of moments around point C:

We can set up the equation as follows:

• The distance $d$ can be calculated as:
  $d = 90 - x$
  where $x$ is the distance from A to C.

From equilibrium of moments, we know:

$140 \cdot R = 210 \cdot d \Rightarrow 140 \cdot 70 = 210 \cdot d.$

On solving, this leads us to find: $d = 60\, cm,$

Since $AB$ is given as $d + x$, where $x$ is the other segment: $$AB = 60 + 60 = 120 \text{ cm}.$

With the new condition that the tension in the cable at C is now three times the tension in the cable at A:

Let the tension at A be $S$. Then the tension at C is $3S$. Summing up the forces:

Using the equilibrium of forces: $210 + W = 3S.$

Also, using the moments about A: $0 = W \cdot (120) - 3S \cdot (90).$

Substituting $W = 30$ into the equation: $W = 3S = 210 + W \Rightarrow 30 = 30.$