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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 (CM) of the non-uniform beam AD, we can apply the principle of moments. Taking moments about point A, we have:
egin{align*}
T + 2T imes (2 - 3) = W
\Rightarrow T - 6T = W
\Rightarrow T(1 - 6) = W
\Rightarrow T = \frac{W}{5}
\end{align*}
The distance of the center of mass d from A can be evaluated using:
d = \frac{2}{3} (4) = \frac{8}{3} extrm{ m}
Thus, the final distance of the center of mass from A is: (\frac{8}{3} extrm{ m}).
Step 2
an expression for the tension in the rope attached to B, giving your answer in terms of k and W.
Answer
Using equilibrium conditions for the load at D, we can write:
egin{align*}
\text{Let } T_B = \text{tension in rope B, and } T_C = \text{tension in rope C.}
\Rightarrow T_B + 2T_B = W + kW
\Rightarrow 3T_B = (1 + k)W
\Rightarrow T_B = \frac{(1 + k)W}{3}
\end{align*}
This gives us the expression for the tension in the rope attached to B: ( T_B = \frac{(1 + k)W}{3} ).
Step 3
the set of possible values of k for which both ropes remain taut.
Answer
To ensure that both tensions are positive, we establish the following inequalities:
-
For T_B ≥ 0, we have: ( (1 + k)W \geq 0 ) Thus, ( k \geq -1 ).
-
For T_C > 0, we evaluate: ( 3T_B - (k - 3) > 0 ) Solving gives us the inequality: ( k < \frac{2}{3} )
Combining these conditions, the possible values of k are: ( 0 < k \leq \frac{2}{3} ) or ( 0 < k < 2/3 ).