a uniform rod,1m long weighing 100N is supported horizontally on two knife edges placed 10cm from its end what will be the reactions at the support when a 40N load is suspended 10cm from the mid point of the rod
To determine the reactions at the supports, we need to analyze the equilibrium of the rod.
Let's consider the left side of the rod first:
Weight of the rod = 100N
Support distance from the left end = 10cm
Since the rod is uniform, we can assume the weight acts through the center of the rod. The weight can be split into two equal parts, each acting at a distance of 50cm from the left end.
So, the left side reaction (R1) at the support can be calculated using the principle of moments:
Moment of the weight about the left support = moment of R1 about the left support
(50cm) * (100N) = (10cm) * (R1)
5000 Ncm = 10 R1
R1 = 500 N
Now, let's consider the right side of the rod:
Load = 40N
Load distance from the midpoint of the rod = 10cm
The load also acts through the center of the rod. The load can be split into two equal parts, each acting at a distance of 50cm from the midpoint.
So, the right side reaction (R2) at the support can be calculated using the principle of moments:
Moment of the load about the right support = moment of R2 about the right support
(50cm) * (40N) = (10cm) * (R2)
2000 Ncm = 10 R2
R2 = 200 N
Therefore, the reactions at the supports are:
R1 = 500 N (left support)
R2 = 200 N (right support)