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A uniform rod, [AB], of length 4 m and weight 80 N is smoothly hinged at end A to a horizontal floor - Leaving Cert Applied Maths - Question 7 - 2012

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A uniform rod, [AB], of length 4 m and weight 80 N is smoothly hinged at end A to a horizontal floor. One end of a light inelastic string is attached to B and the o... show full transcript

Worked Solution & Example Answer:A uniform rod, [AB], of length 4 m and weight 80 N is smoothly hinged at end A to a horizontal floor - Leaving Cert Applied Maths - Question 7 - 2012

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:

  • The weight of the rod (80 N) acting downwards at its center of gravity (which is at the midpoint of the rod).
  • The tension in the string (T), acting at point B, directed along the string at an angle of 60° to the ceiling.
  • The reaction at point A, which can be broken into horizontal (X) and vertical (Y) components.

Step 2

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

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Answer

Resolving the forces horizontally:

X=Timesextcos(60°)X = T imes ext{cos}(60°)

Resolving the forces vertically:

Y+Timesextsin(60°)=80Y + T imes ext{sin}(60°) = 80

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 gives:

Times4=80imes2imesextcos(30°)T imes 4 = 80 imes 2 imes ext{cos}(30°)

Step 4

Find the tension in the string.

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Answer

From the moments equation:

Times4=80imes2imesextcos(30°)T imes 4 = 80 imes 2 imes ext{cos}(30°)

Solving for T:

T = rac{80 imes 2 imes ext{cos}(30°)}{4} = 20 imes rac{ ext{sqr}(3)}{2} = 20 ext{/3}

Step 5

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

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Answer

To find the reaction at hinge A, we can use:

R = ext{sqr}igg{(}igg{(}T imes ext{cos}(60°)igg{)}^2 + igg{(}Y - 80 + T imes ext{sin}(60°)igg{)}^2igg{)}

Substituting the found values yields:

R = rac{ ext{sqr}igg{(}(0 + 50igg{)}^2 + (80 - T imes ext{sin}(60°))^2}{20/ ext{sqr}(3)}}

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