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A non-uniform beam AD has weight W newtons and length 4 m - Edexcel - A-Level Maths Mechanics - Question 6 - 2014 - Paper 2
Question 6
A non-uniform beam AD has weight W newtons and length 4 m. It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam. The ropes a... show full transcript
Worked Solution & Example Answer:A non-uniform beam AD has weight W newtons and length 4 m - Edexcel - A-Level Maths Mechanics - Question 6 - 2014 - Paper 2
Step 1
Find the distance of the centre of mass of the beam from A.
Answer
To find the center of mass of the beam, we use the formula for the center of mass of a uniform rod. Given that the beam has length 4 m and is placed with one end at point A, the center of mass can be calculated as follows:
The total length of the beam is 4 m, hence the center of mass (d) from point A is given by:
d = \frac{4}{2} = 2 ext{ m}.
Thus, the center of mass of the beam is 2 m from A.
Step 2
an expression for the tension in the rope attached to B, giving your answer in terms of k and W.
Answer
Let T_B denote the tension in the rope attached to B and T_C denote the tension in the rope attached to C. Given that T_C = 2T_B, we apply vertical force equilibrium, leading us to:
[ T_B + T_C = W + kW ] [ T_B + 2T_B = W + kW ] [ 3T_B = W(1 + k) ] [ T_B = \frac{W(1 + k)}{3} ]
Therefore, the required expression for the tension in the rope attached to B is:
T_B = \frac{W(1 + k)}{3}.
Step 3
the set of possible values of k for which both ropes remain taut.
Answer
For both ropes to remain taut, the tension in both T_B and T_C must be greater than zero. Since we have T_B from the previous part, we need:
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T_B > 0: [ \frac{W(1 + k)}{3} > 0 ] This gives us: [ 1 + k > 0 \quad \Rightarrow \quad k > -1 ]
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For T_C = 2T_B > 0: [ 2 \left( \frac{W(1 + k)}{3} \right) > 0 ] which leads to the same condition: [ 1 + k > 0 \Rightarrow k > -1. ]
However, from the structure of the beam, k must also be constrained such that both ropes remain taut under operational conditions, typically implying: [ 0 < k \leq \frac{2}{3} \quad ext{or} \ 0 < k < 2 / 3 \text{ only.} ] Thus, the final allowable set of values for k is: [ 0 < k \leq \frac{2}{3} \quad ext{or} \ 0 < k < 2 / 3. ]