A uniform rod, AB, of length 4 m and weight 160 N is smoothly hinged at end A to a horizontal ceiling. One end of a light inextensible string is attached to B and the other end of the string is attached to a vertical wall. The rod makes an angle of 60° with the ceiling and the string makes an angle of 60° with the wall, as shown in the diagram. The rod is in equilibrium. (i) Show on a diagram all the forces acting on the rod AB. (ii) Write down the two equations that arise from resolving the forces horizontally and vertically. (iii) Write down the equation that arises from taking moments about the point A. (iv) Find the tension in the string. (v) Find the magnitude of the reaction at the point A.

Leaving Cert Applied Maths Statics: A uniform rod, AB, of length 4 m

A uniform rod, AB, of length 4 m and weight 160 N is smoothly hinged at end A to a horizontal ceiling - Leaving Cert Applied Maths - Question 7 - 2016

Question 7

A uniform rod, AB, of length 4 m and weight 160 N is smoothly hinged at end A to a horizontal ceiling. One end of a light inextensible string is attached to B and t... show full transcript

Worked Solution & Example Answer:A uniform rod, AB, of length 4 m and weight 160 N is smoothly hinged at end A to a horizontal ceiling - Leaving Cert Applied Maths - Question 7 - 2016

Step 1

Show on a diagram all the forces acting on the rod AB.

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Answer

In the diagram, we have the following forces acting on the rod AB:

The weight of the rod (W) acting downwards at the center (point O) which is 160 N.
The tension in the string (T) acting at point B, directed diagonally towards the wall.
The reaction forces at point A, which can be broken into two components: horizontal (X) and vertical (Y) components.

The forces can be summarized as follows:

Weight (W = 160 N) acting vertically downwards.
Tension (T) making an angle of 60° with the rod.
Reaction at A with components X (horizontal) and Y (vertical).

Step 2

Write down the two equations that arise from resolving the forces horizontally and vertically.

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Answer

From resolving the forces horizontally and vertically, we obtain the following equations:

Horizontally: $T \, \cos(30°) = X$
Vertically: $T \, \sin(30°) + Y = 160$

Step 3

Write down the equation that arises from taking moments about the point A.

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Answer

Taking moments about point A, we have:

$T \, \cdot 4 \times \cos(60°) = 160 \times 2$

This simplifies to:

$4T \cos(60°) = 160 \times 2$

Step 4

Find the tension in the string.

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Answer

Using the equation from taking moments, we substitute (\cos(60°) = \frac{1}{2}):

$4T \cdot \frac{1}{2} = 320$

Solving for T, we get:

$T = \frac{320}{2 \cdot 4} = 40 \, N$

Step 5

Find the magnitude of the reaction at the point A.

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Answer

Using the previously determined values:

From the horizontal equilibrium: $X = 40 \cdot \frac{\sqrt{3}}{2} = 20\sqrt{3}$
From the vertical equilibrium: $Y = 160 - 40 \cdot \frac{1}{2} = 140$

We can then calculate the magnitude of the reaction at point A as follows:

$R = \sqrt{(20\sqrt{3})^2 + 140^2}$

This computes to:

$R = \sqrt{(400 \cdot 3) + 19600} = \sqrt{1200 + 19600} = \sqrt{20800} = 144.2 \, N$

A uniform rod, AB, of length 4 m and weight 160 N is smoothly hinged at end A to a horizontal ceiling - Leaving Cert Applied Maths - Question 7 - 2016

Worked Solution & Example Answer:A uniform rod, AB, of length 4 m and weight 160 N is smoothly hinged at end A to a horizontal ceiling - Leaving Cert Applied Maths - Question 7 - 2016

Show on a diagram all the forces acting on the rod AB.

Write down the two equations that arise from resolving the forces horizontally and vertically.

Write down the equation that arises from taking moments about the point A.

Find the tension in the string.

Find the magnitude of the reaction at the point A.

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