A uniform beam 6m long and weight 4kg rest on support p and q place left and right 1.0m from each end of the beam weight of mass 10kgand 8kg are place near p and q respectively on each end of the beam.calculate the reaction p and q
draw the picture
calculate the moments using p, and then q as the pivot
As a student seaking for help
To calculate the reactions at supports P and Q, we need to analyze the forces acting on the beam.
Let's consider the beam as a whole system. The weight of the beam itself (4 kg) can be considered as acting at its center of mass, which is at a distance of 3 m from either end.
First, let's calculate the total weight acting at the center of the beam:
Weight of the beam = 4 kg × 9.8 m/s^2 (acceleration due to gravity) = 39.2 N
Now, let's analyze the forces acting on the beam section near support P:
1. Weight of the beam: 19.6 N (half of the total weight since it's equidistant from both ends)
2. Weight of the 10 kg mass: 10 kg × 9.8 m/s^2 = 98 N
Therefore, the total downward force acting near support P is:
Force at P = 19.6 N + 98 N = 117.6 N
Next, let's analyze the forces acting on the beam section near support Q:
1. Weight of the beam: 19.6 N (half of the total weight since it's equidistant from both ends)
2. Weight of the 8 kg mass: 8 kg × 9.8 m/s^2 = 78.4 N
Therefore, the total downward force acting near support Q is:
Force at Q = 19.6 N + 78.4 N = 98 N
Finally, to calculate the reactions at supports P and Q, we need to consider that the beam is in equilibrium, meaning that the sum of all the vertical forces acting on it should be zero. Since the forces acting downwards are greater than the upward forces, we can assume the reactions at P and Q are upward forces.
The reaction at support P (RP) is equal to the total downward force acting near support P:
RP = 117.6 N
The reaction at support Q (RQ) is equal to the total downward force acting near support Q:
RQ = 98 N
Therefore, the reaction at support P is 117.6 N upward, and the reaction at support Q is 98 N upward.