A uniform rod (AB) of mass 80kg and 1.0m lenght is supported on the 2 knife edges placed 15.0cm from its ends. what will be the reactions at these support when a 20kg mass is suspend 10cm from the mid-point of the rod.(workings and diagram).
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To solve this problem, we need to calculate the reactions at the two support points on the rod.
First, let's draw a diagram to understand the scenario better:
```
<-----15.0cm-----><-----50.0cm-----><-----15.0cm----->
|_________________________________________|
^ ^
| C |
| |
| |
Rod Mass (20kg)
```
Here, the rod AB is 1.0m long. The two knife edges are placed at 15.0cm from each end, and the mass of the rod is 80kg. The 20kg mass is suspended 10cm from the mid-point of the rod (at point C).
To find the reaction forces at the two support points, we can apply the principle of moments. The total anticlockwise moments must be balanced by the total clockwise moments.
1. Calculating the moments due to the rod's weight:
The weight acts vertically downward at the center of the rod, so its moment about the support at A is zero since it passes through the point of rotation.
2. Calculating the moments due to the suspended mass at point C:
The weight of the 20kg mass generates moments in both the clockwise and anticlockwise directions. The moment generated by the mass about the support at A is given by:
Moment(A) = Weight(C) * Distance(AC)
= (20kg * 9.8m/s^2) * (0.15m - 0.10m) [using the distance from point A to C]
3. Applying the principle of moments:
The sum of the clockwise moments must equal the sum of the anticlockwise moments. Since the moment generated by the rod is zero, we have:
(Reactions at A * 0.15m) = Moment(A)
4. Calculating the reactions at the support points:
Reactions at A = Moment(A) / 0.15m
Now, let's calculate the reactions at the support points:
Weight(C) = mass(C) * acceleration due to gravity = 20kg * 9.8m/s^2 = 196N
Distance(AC) = 0.15m - 0.10m = 0.05m
Moment(A) = (20kg * 9.8m/s^2) * (0.05m) = 98N.m
Reactions at A = Moment(A) / 0.15m = 98N.m / 0.15m = 653.33N
Therefore, the reaction at each support point will be approximately 653.33N.