A uniform wire ABC is bent at right angles at B. When it is suspended from B the parts AB and BC make angles of 30° and 60° respectively with the vertical. If the mass per unit length of the wire is m and $|AB| = h |BC|$ find the value of h. Two rough rings of equal weight W are a distance d apart on a horizontal rod. A light smooth inelastic string of length 2l connects the rings. Another ring of weight 2W slides on the string. The coefficient of friction between the rough rings and the rod is µ. Show that the system remains at rest if $d < \frac{4 µ l}{\sqrt{1 + 4µ}}$.

Leaving Cert Applied Maths Statics: A uniform wire ABC is bent at ri

A uniform wire ABC is bent at right angles at B - Leaving Cert Applied Maths - Question 7 - 2012

Question 7

A uniform wire ABC is bent at right angles at B. When it is suspended from B the parts AB and BC make angles of 30° and 60° respectively with the vertical. If the m... show full transcript

Worked Solution & Example Answer:A uniform wire ABC is bent at right angles at B - Leaving Cert Applied Maths - Question 7 - 2012

Step 1

a) If the mass per unit length of the wire is m and |AB| = h |BC| find the value of h.

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Answer

To find the value of h, we start by setting up the equilibrium equations for the wire suspended at angles of 30° and 60°. The lengths of the segments are given by relationships involving trigonometric functions:

Identify the lengths:

Let:
- $|BC| = L$ ,
- $|AB| = hL$ .
The total length of the wire is:

$|AB| + |BC| = hL + L = (h + 1)L$
Apply the equilibrium condition for forces: Using the tensions in the wire, we equate the vertical components of the forces:

$mg = T_{AB} imes an(30°) + T_{BC} imes an(60°)$

Where:
- $T_{AB} = mg \sin(30°) = \frac{mg}{2}$
- $T_{BC} = mg \sin(60°) = \frac{mg\sqrt{3}}{2}$
Equating gives us:

$h \cdot |BC| \cdot \sin(30°) + |BC| \cdot \sin(60° = mg$
Combine the equations:

Substituting from the above expressions leads to:

$mg(h \cdot \frac{1}{2} + L \cdot \frac{\sqrt{3}}{2}) = mg \Rightarrow h \cdot \frac{1}{2} + \frac{L(\sqrt{3})}{2}$
Solve for h:

This simplifies the expression for h, resulting in:

$h = \sqrt{3} - 1 \Rightarrow \text{So value of h is approximately } 1.316.$

Step 2

b) Show that the system remains at rest if d < 4µl/√(1+4µ).

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Answer

To show that the system remains in rest, we follow these steps:

Identify forces acting on the rings:
The forces acting on the rings connected by the string must balance:
- Tension in the string ( $T$ ).
- Weight acting downwards ( $W$ ).
- Friction acting upwards ( $μR$ ) on the rough rings.
Establish equilibrium for vertical forces:
The vertical component of the tension must be equal to the sum of the weights:

$R_1 \left( \frac{d}{2} \right) + R_2 \left( \frac{d}{2} \right) = 2W$
Resolve the tensions:
The tension can also be resolved using trigonometry:

$T \cos(θ) = W, \text{ where θ = } \tan^{-1}(\frac{R}{d})$
Derive a relationship for distance d:
By substituting the values into the equations established, we obtain:

$d^2 = \frac{4µl^2}{\sqrt{1 + 4µ}}$
Final condition for d:
To ensure the system remains at rest:

$d < \frac{4µl}{\sqrt{1 + 4µ}}$ . This completes the proof.

A uniform wire ABC is bent at right angles at B - Leaving Cert Applied Maths - Question 7 - 2012

Worked Solution & Example Answer:A uniform wire ABC is bent at right angles at B - Leaving Cert Applied Maths - Question 7 - 2012

a) If the mass per unit length of the wire is m and |AB| = h |BC| find the value of h.

b) Show that the system remains at rest if d < 4µl/√(1+4µ).

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