A pole AB has length 3 m and weight W newtons - Edexcel - A-Level Maths Mechanics - Question 4 - 2010 - Paper 1

Let the tension in the rope at A be ( X ). From the equilibrium of forces in the vertical direction:

$$X + Y = W + 20.$$

Substituting for ( Y ) from part (a):

$$X + \left( \frac{5}{6} W + \frac{100}{3} \right) = W + 20.$$

Simplifying this, we find:

$$X = W + 20 - \left( \frac{5}{6} W + \frac{100}{3} \right) = \frac{1}{6} W - \frac{40}{3}.$$

Using the condition that the tension in the rope attached to the pole at C is eight times the tension in the rope attached to the pole at A, we have:

$$Y = 8X.$$

Substituting the expressions for ( X ) and ( Y ):

$$\frac{5}{6} W + \frac{100}{3} = 8 \left( \frac{1}{6} W - \frac{40}{3} \right).$$

Solving this equation:

1. Distributing on the right gives:

$$\frac{5}{6} W + \frac{100}{3} = \frac{8}{6} W - \frac{320}{3}.$$

2. Now combine like terms:

$$\frac{5}{6} W - \frac{8}{6} W = -\frac{320}{3} - \frac{100}{3}.$$

3. This simplifies to:

$$- \frac{3}{6} W = -\frac{420}{3}.$$

4. Thus:

$$W = 280.$$