A uniform rod 1 metre long weighing 50N is supported horizontally on two knife edges placed 10cm each from its ends. What will be the reactions at these supports when 100N weight is suspended 10cm from the midpoint of the rod?
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A uniform rod 1m long weighing 50N is supported on horizontal two knives edges placed 10cm each from it's end.What will be the reaction at this support when a 100N weight is suspended 10cm from the mid point of the rod.
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To find the reactions at the supports, we need to analyze the equilibrium of the rod.
First, let's consider the forces acting on the rod:
1. Weight of the rod (acting at the center):
The weight of the rod is given as 50N and it acts at the center of the rod.
This force can be represented as a single downward force acting at the midpoint.
2. Reaction forces at the supports:
Let's denote the reaction force at the left support as R1 and the reaction force at the right support as R2.
3. Weight suspended from the midpoint:
The weight of 100N suspended 10cm (i.e., 0.1m) from the midpoint of the rod.
Now, to find the reactions at the supports, we need to apply the principle of moments (also known as torque) about any point.
Let's take the left support as the point of moment:
Sum of anticlockwise moments = Sum of clockwise moments
The moments can be calculated by multiplying the force by its perpendicular distance from the point of moment.
For the anticlockwise moments:
Moment due to weight of the rod = 50N * (0.10m + 0.10m) (distance from the center to the left support + distance from the center to the right support)
For the clockwise moments:
Moment due to weight suspended from the midpoint = 100N * 0.10m (distance from the left support to the weight)
Since the rod is in equilibrium, the sum of anticlockwise moments should be equal to the sum of clockwise moments.
Therefore, (50N * 0.20m) + (R1 * 0.10m) = (100N * 0.10m)
Now, we can solve this equation to find the reaction force at the left support (R1).