A pole AB has length 3 m and weight W newtons. The pole is held in a horizontal position in equilibrium by two vertical ropes attached to the pole at the points A and C where AC = 1.8 m, as shown in Figure 2. A load of weight 20 N is attached to the rod at B. The pole is modelled as a uniform rod, the ropes as light inextensible strings and the load as a particle. (a) Show that the tension in the rope attached to the pole at C is $\frac{5}{6}W + \frac{100}{3} \text{ N}$. (b) Find, in terms of W, the tension in the rope attached to the pole at A. (c) Find the value of W.

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

Question 4

A pole AB has length 3 m and weight W newtons. The pole is held in a horizontal position in equilibrium by two vertical ropes attached to the pole at the points A an... show full transcript

Worked Solution & Example Answer:A pole AB has length 3 m and weight W newtons - Edexcel - A-Level Maths Mechanics - Question 4 - 2010 - Paper 1

Step 1

Show that the tension in the rope attached to the pole at C is $\frac{5}{6}W + \frac{100}{3}$ N.

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Answer

To find the tension in the rope at point C, we can consider the moments about point A. Setting the sum of the moments equal to zero for equilibrium, we have:

$W \times 1.5 + 20 \times 3 = Y \times 1.8$

Substituting the expressions gives us:

$Y = \frac{W \times 1.5 + 20 \times 3}{1.8}$

This can be simplified:

$Y = \frac{5}{6}W + \frac{100}{3}$

Thus, we have shown the required result.

Step 2

Find, in terms of W, the tension in the rope attached to the pole at A.

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Answer

Let the tension in the rope at A be denoted as X. From the equilibrium of forces in the vertical direction, we can express:

$X + Y = W + 20$

Substituting for Y from the previous calculation:

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

Solving for X:

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

This can be simplified to provide the expression for X in terms of W.

Step 3

Find the value of W.

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Answer

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

$Y = 8X$

Substituting the expressions derived for Y and X leads us to:

$\frac{5}{6}W + \frac{100}{3} = 8 \left(W + 20 - \left(\frac{5}{6}W + \frac{100}{3}\right)\right)$

This equation can then be solved to find the value of W. After simplification and calculation, we find:

$W = 280$ .

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

Worked Solution & Example Answer:A pole AB has length 3 m and weight W newtons - Edexcel - A-Level Maths Mechanics - Question 4 - 2010 - Paper 1

Show that the tension in the rope attached to the pole at C is \(\frac{5}{6}W + \frac{100}{3}\) N.

Find, in terms of W, the tension in the rope attached to the pole at A.

Find the value of W.

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