Aplank AB 3.0m long weighing 20kg and with it centre of gravity 2.0m from the end A.It rest on two support at Cand D. COMPUTE THE VALUES OF THE REACTION

Answer 1

To solve for the reactions at supports C and D for the plank, we can use the principles of equilibrium. A plank in static equilibrium will satisfy two conditions:

1. The sum of vertical forces must equal zero.
2. The sum of moments around any point must equal zero.

Given:

• Length of the plank (L) = 3.0 m
• Weight of the plank (W) = 20 kg
• Distance of the center of gravity from point A (x) = 2.0 m
• Weight in Newtons (W) = mass * gravitational acceleration = 20 kg * 9.81 m/s² = 196.2 N

Steps:

1. Identify the positions of the supports and the forces acting on the beam:

   • Plank is supported at points C and D.
   • Let \( R_C \) be the reaction force at support C, and \( R_D \) be the reaction force at support D.
   • The weight of the plank acts vertically downward at its center of gravity, which is 2.0 m from point A.

2. Position of Supports:

   • Assuming support C is at point A (0 m) and support D is at point B (3 m):
     • Center of gravity is at point (2.0 m) from A or (1.0m) from D.

3. Set up equations for equilibrium:

• Sum of Vertical Forces = 0:

\[ R_C + R_D - W = 0 \ R_C + R_D - 196.2 = 0 \tag{1} \]

• Sum of Moments about point A = 0:

Taking moments about point A (moment = force × distance):

For counter-clockwise moments as positive:

\[ R_D \cdot 3.0 - W \cdot 2.0 = 0 \ R_D \cdot 3.0 - 196.2 \cdot 2.0 = 0 \]

Solving for \( R_D \):

\[ R_D \cdot 3.0 = 392.4 \ R_D = \frac{392.4}{3.0} = 130.8 , N \tag{2} \]

4. Substituting \( R_D \) in equation (1):

\[ R_C + R_D = 196.2 \ R_C + 130.8 = 196.2 \ R_C = 196.2 - 130.8 = 65.4 , N \tag{3} \]

Final Reactions

• The reaction at support C, \( R_C \) = 65.4 N
• The reaction at support D, \( R_D \) = 130.8 N

So, \( R_C = 65.4 , N \) (upward) and \( R_D = 130.8 , N \) (upward).