A uniform beam 6.0metres long and weighing 4kg rests on support p and q placed left and rights 1.0metre from each end of the beam .weights of mass 10kg and 8kg are placed near p and q respectively on each end of the beam . calculate the reactions at p and q
We can first calculate the total weight of the beam and the weights:
Total weight = (weight of beam) + (weight on left end) + (weight on right end)
Total weight = 4kg + 10kg + 8kg
Total weight = 22kg
Since the beam is in equilibrium, the sum of the forces in the vertical direction must be equal to zero.
Let R_p and R_q be the reactions at supports P and Q, respectively.
∑F_y = 0
R_p + R_q - 22kg = 0
We can also take moments about support P (or Q) to obtain another equation.
Taking moments about P:
∑M_p = 0
(R_q x 5m) - (10kg x 1m) - (4kg x 3m) = 0
Simplifying:
R_q = (10kg x 1m) + (4kg x 3m) / 5m
R_q = 6.8kg
Substituting this value of R_q into the first equation:
R_p + 6.8kg - 22kg = 0
R_p = 15.2kg
Therefore, the reaction at support P is 15.2kg and the reaction at support Q is 6.8kg.