A uniform rod, [AB], of length 2 m and weight 40 N is smoothly hinged at end A to a vertical wall. One end of a light inelastic string is attached to B and the other end of the string is attached to a horizontal ceiling. The string makes an angle of 30° with the ceiling and the rod makes an angle of 30° 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 point A. (iv) Find the tension in the string. (v) Find the magnitude of the reaction at the hinge, A.

Leaving Cert Applied Maths Statics: A uniform rod, [AB], of length 2

A uniform rod, [AB], of length 2 m and weight 40 N is smoothly hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2010

Question 7

A uniform rod, [AB], of length 2 m and weight 40 N is smoothly hinged at end A to a vertical wall. One end of a light inelastic string is attached to B and the othe... show full transcript

Worked Solution & Example Answer:A uniform rod, [AB], of length 2 m and weight 40 N is smoothly hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2010

Step 1

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

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Answer

The forces acting on the rod [AB] include:

The weight of the rod (40 N) acting downwards at its midpoint.
The tension (T) in the string acting at point B, directed at an angle of 30° to the horizontal.
The horizontal reaction force (X) at hinge A, directed towards the wall.
The vertical reaction force (Y) at hinge A, acting upwards.

Step 2

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

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Answer

Resolving forces horizontally: $X = T \cos 30°$

Resolving forces vertically: $Y + T \sin 30° = 40$

Step 3

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

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Answer

Taking moments about point A: $T(2) = 40 \times (1)$

This simplifies to: $T(2) = 40 \sin 30°$

Step 4

Find the tension in the string.

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Answer

From the moment equation: $T(2) = 40 \sin 30°$ Substituting the value of \sin 30° = 0.5: $T(2) = 40 \times 0.5\Rightarrow T = 10 \text{ N}$

Step 5

Find the magnitude of the reaction at the hinge, A.

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Answer

Using the equations from horizontal and vertical resolutions:

Substituting for T: $X = T \cos 30° = 10 \cos 30° = 5 \sqrt{3}$
From the vertical equation: $Y + 10 \sin 30° = 40 \Rightarrow Y + 10 \times 0.5 = 40 \Rightarrow Y = 35$

Finally, the magnitude of the reaction at hinge A: $R = \sqrt{(5 \sqrt{3})^2 + 35^2}$ Calculating: $R = \sqrt{(75) + (1225)} = \sqrt{1300} \approx 36.1 \text{ N}$

A uniform rod, [AB], of length 2 m and weight 40 N is smoothly hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2010

Worked Solution & Example Answer:A uniform rod, [AB], of length 2 m and weight 40 N is smoothly hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2010

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 point A.

Find the tension in the string.

Find the magnitude of the reaction at the hinge, A.

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