İ) What is the magnitude and the direction of the total moment with respect to perpendicular line passing through point O? ii) Find the reaction force at pivot point O

Eğitim ve Öğretim Matematik
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(e) Weightless T shaped rod which is pivoted on point O and the forces acting on it are on the same plane.

i) What is the magnitude and the direction of the total moment with respect to perpendicular line passing through point O?

ii) Find the reaction force at pivot point O.

Resim, bir duvara sabitlenmiş olan ağırlıksız bir T şeklindeki çubuğun üzerindeki kuvvetleri ve momentleri gösteren bir mekanik problemi ifade ediyor. Problemin parçaları ve sorulan sorular yazılı olarak verilmiş. Çubuğun üst kısmında sağa doğru 10 N’lik bir kuvvet ve aynı kolda sola doğru 40 N’lik bir kuvvet uygulanmakta. Dikey kolda aşağı yukarı yönlü iki 40 N’lik kuvvet bulunmakta. Soruda çubuğun momentinin büyüklüğü ve yönü ile O pivot noktasındaki reaksiyon kuvveti sorulmaktadır.

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For the T-shaped rod in question, we can solve for both the total moment about point O and the reaction force at that pivot point.

Part i) Total Moment Calculation

We define the counterclockwise direction as positive for moments. We calculate the moment about point O due to each force:

  1. Force at Top (10 N):

    • Distance from O: 2 m (1 m right + 1 m up)
    • Moment = Force x Distance = 10 \, \text{N} \times 2 \, \text{m} = 20 \, \text{Nm} (clockwise, thus negative)

  2. Force at Right (40 N):

    • Distance from O: 1 m (right arm)
    • Moment = 40 \, \text{N} \times 1 \, \text{m} = 40 \, \text{Nm} (counterclockwise, positive)

  3. Force at Bottom (40 N):

    • Distance from O: 1 m (left extension)
    • Moment = 40 \, \text{N} \times 1 \, \text{m} = 40 \, \text{Nm} (counterclockwise, positive)

Adding these moments:
[ \text{Total Moment} = -20 , \text{Nm} + 40 , \text{Nm} + 40 , \text{Nm} = 60 , \text{Nm} ]
Thus, the total moment about point O is 60 Nm in the counterclockwise direction.

Part ii) Reaction Force at Pivot Point O

The reaction at O must counteract all the external forces to maintain static equilibrium. Here are the vertical and horizontal components of the external forces:

  • Horizontal:

    • 40 N to the right cancels with 40 N to the left.
    • Net horizontal force = 40 \, \text{N} - 40 \, \text{N} = 0 \, \text{N}

  • Vertical:

    • 10 N upwards from the top end.
    • Net vertical force = 10 \, \text{N} upwards (no downward external forces).

So, the reaction at pivot point O needs to provide:

  • Horizontally: 0 N (no horizontal reaction is necessary as forces balance out).
  • Vertically: 10 N downwards to counteract the 10 N upward force.

Therefore, the reaction force at O is 10 N directed downwards.