A uniform rod, [ab], of length 4 m and weight 100 N is smoothly hinged at end a to a horizontal floor. One end of a light inelastic string is attached to b and the other end of the string is attached to a vertical wall. The string makes an angle of 60° with the wall and the rod makes an angle of 60° with the floor, 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.

Leaving Cert Applied Maths Statics: A uniform rod, [ab], of length 4

A uniform rod, [ab], of length 4 m and weight 100 N is smoothly hinged at end a to a horizontal floor - Leaving Cert Applied Maths - Question 7 - 2008

Question 7

A uniform rod, [ab], of length 4 m and weight 100 N is smoothly hinged at end a to a horizontal floor. One end of a light inelastic string is attached to b and the ... show full transcript

Worked Solution & Example Answer:A uniform rod, [ab], of length 4 m and weight 100 N is smoothly hinged at end a to a horizontal floor - Leaving Cert Applied Maths - Question 7 - 2008

Step 1

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

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Answer

In the diagram, we represent the forces acting on the rod [ab] as follows:

Weight of the rod (W) acting downwards at the midpoint, which has a magnitude of 100 N.
Tension (T) in the string acting at an angle of 60° from the wall.
Reaction force (R) at the hinge (point a), acting vertically upwards and horizontally towards the rod.

Step 2

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

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Answer

For horizontal forces:

$T imes ext{sin}(60°) = X$

For vertical forces:

$Y + T imes ext{cos}(60°) = 100$

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 the equation:

$T(4) = 100 imes (2 imes ext{cos}(60°))$

Step 4

Find the tension in the string.

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Answer

From the moment equation, we solve for T:

$T = \frac{100 imes (2 imes ext{cos}(60°))}{4} = 25 ext{ N}$

Step 5

Find the magnitude of the reaction at the hinge.

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Answer

First calculate the vertical reaction:

$Y + 25 imes ext{cos}(60°) = 100$

Thus,

$Y = 100 - 25 imes 0.5 = 87.5 ext{ N}$

For horizontal reaction:

$X = 25 imes ext{sin}(60°) = 21.65 ext{ N}$

The resultant reaction at the hinge is:

$\sqrt{X^2 + Y^2} = \sqrt{(21.65)^2 + (87.5)^2} = 90.1 ext{ N}$

A uniform rod, [ab], of length 4 m and weight 100 N is smoothly hinged at end a to a horizontal floor - Leaving Cert Applied Maths - Question 7 - 2008

Worked Solution & Example Answer:A uniform rod, [ab], of length 4 m and weight 100 N is smoothly hinged at end a to a horizontal floor - Leaving Cert Applied Maths - Question 7 - 2008

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.

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