A uniform rod, AB, of length 2 m and weight 40√3 N is freely 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 another vertical wall, as shown in the diagram. The rod makes an angle of 60° with one wall and the string makes an angle of 30° with the other 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 point A.

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

A uniform rod, AB, of length 2 m and weight 40√3 N is freely hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2018

Question 7

A uniform rod, AB, of length 2 m and weight 40√3 N is freely hinged at end A to a vertical wall. One end of a light inelastic string is attached to B and the other ... show full transcript

Worked Solution & Example Answer:A uniform rod, AB, of length 2 m and weight 40√3 N is freely hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2018

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 are as follows:

The weight of the rod acting vertically downward at the midpoint, which is 40√3 N.
The tension T in the string pulling at an angle of 30° from the horizontal at point B.
The reaction force R acting at point A, which can be resolved into two 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

For horizontal forces: $X = T \cos(60°)$

For vertical forces: $40\sqrt{3} = Y + T \sin(60°)$

Step 3

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

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Answer

Taking moments about point A gives us: $T \times 2 = 40\sqrt{3} \times \frac{\sqrt{3}}{2}$ This simplifies to: $2T = 40\sqrt{3} \times \frac{\sqrt{3}}{2}$

Step 4

Find the tension in the string.

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Answer

From the moment equation: $2T = 30\sqrt{3}$ Dividing both sides by 2: $T = 15N$

Step 5

Find the magnitude of the reaction at the point A.

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Answer

Using the horizontal force equation: $X = T \cos(60°) = 15$

Using the vertical force equation: $Y = 40\sqrt{3} - 15 \sin(60°)$

after calculation: $Y = 25\sqrt{3}$

The reaction R at point A can be found using Pythagorean theorem: $R = \sqrt{X^2 + Y^2} = \sqrt{15^2 + (25\sqrt{3})^2}$ Calculating gives: $R = 45.8 N$

A uniform rod, AB, of length 2 m and weight 40√3 N is freely hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2018

Worked Solution & Example Answer:A uniform rod, AB, of length 2 m and weight 40√3 N is freely hinged at end A to a vertical wall - Leaving Cert Applied Maths - Question 7 - 2018

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 point A.

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