A light inextensible string has one end, A, attached to a fixed point and
the other end, B to a small, smooth ring of negligible weight. Another string,
CBD, has one end attached to a fixed point C on the same horizontal level as
A, passes through the ring at B, and supports a hanging weight, W, at its other
end. In equilibrium AB is inclined at 65
to the vertical. Find the tension in
AB?
I cannot seem to do this question, what should the diagram look like? I'm so confused please help, 10 points to best answer!
the other end, B to a small, smooth ring of negligible weight. Another string,
CBD, has one end attached to a fixed point C on the same horizontal level as
A, passes through the ring at B, and supports a hanging weight, W, at its other
end. In equilibrium AB is inclined at 65
to the vertical. Find the tension in
AB?
I cannot seem to do this question, what should the diagram look like? I'm so confused please help, 10 points to best answer!
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The strings form a Y-shape. A and C are the top ends, B is the centre, and the weight W hangs from the tail, D, of the Y.
With a smooth ring, BAC will form an isosceles triangle, so CB will also have an inclination of 65° to the vertical, and therefore it will be at 25° from the horizontal.
The tension in the string from B to C must be the same as that from D to B (the tension must be the same along the whole length of the string), so the force in the section BC = W.
The horizontal component of this force is W sin 25°. If the system is in equilibrium, then every force is balanced by an equal force in the opposite direction.
Therefore the horizontal force from the tension in BC must be balanced by an equal force from the tension in AB. Therefore the tension in AB must equal the tension in BC, which is W, giving an equal horizontal component of -W cos 25° ( minus because it is in the opposite direction).
With a smooth ring, BAC will form an isosceles triangle, so CB will also have an inclination of 65° to the vertical, and therefore it will be at 25° from the horizontal.
The tension in the string from B to C must be the same as that from D to B (the tension must be the same along the whole length of the string), so the force in the section BC = W.
The horizontal component of this force is W sin 25°. If the system is in equilibrium, then every force is balanced by an equal force in the opposite direction.
Therefore the horizontal force from the tension in BC must be balanced by an equal force from the tension in AB. Therefore the tension in AB must equal the tension in BC, which is W, giving an equal horizontal component of -W cos 25° ( minus because it is in the opposite direction).