A uniform wooden beam of length 2m and weight 34N tests on two supports a and b placed at 40cm from either end of the beam two weight of 40N and 50N are suspended at the end of the beam . (i) :Draw adiagram to show the forces acting on the beam. (ii) Calculate the reactions at the supports.
To draw a diagram showing the forces acting on the beam, we need to consider the weight of the beam itself and the weights suspended at the ends.
(i) Diagram of forces acting on the beam:
- Start by drawing a horizontal line to represent the beam.
- Mark the two supports, a and b, on the line where they are placed (40 cm from either end).
- Draw arrows downward at supports a and b to represent the reactions. Label these reaction forces as Ra and Rb, respectively.
- At the left end of the beam, draw an arrow downward to represent the weight of 40N suspended and label it as W1.
- At the right end of the beam, draw an arrow downward to represent the weight of 50N suspended and label it as W2.
- Finally, draw an arrow upward at the midpoint of the beam to represent the weight of the beam itself. Label this arrow as Wbeam.
(ii) Calculating the reactions at the supports:
Let's denote the reactions at supports a and b as Ra and Rb, respectively.
To find the reactions, we need to consider the equilibrium of forces acting on the beam. Since the beam is in equilibrium, the sum of all the vertical forces acting on it must be equal to 0.
Vertical forces downward: Wbeam + W1 + W2
Vertical forces upward: Ra + Rb
According to equilibrium conditions, the sum of the downward forces must be equal to the sum of the upward forces:
Wbeam + W1 + W2 = Ra + Rb
Given that the weight of the beam (Wbeam) is 34N, W1 is 40N, and W2 is 50N, we can substitute these values into the equation:
34N + 40N + 50N = Ra + Rb
Simplifying this equation, we have:
124N = Ra + Rb
Hence, the reactions at supports a and b are equal to 124N each.