A uniform rod of mass 100g has a length of 1m.it is supported horizantally on two knife edges placed 10cm from it end. What will be the reaction at these supports when a 50g mass is suspended 30cm from one of the knife edges?
See question above
pls answer me
A unuform rod of mass 100g has a length a 1m.it is supported horixontally on two knife edges placed 10cm from its end. what will be the reactions at thise supports when a 50g mass is suspended 30cm from one of the knife edges?
Solve the question
my answer is 17/4 and 7/4
To calculate the reactions at the two knife edges, we can use the concepts of torques and equilibrium.
Let's start by calculating the torque (moment) of each force acting on the rod:
1. The weight of the rod: The weight acts at the center of mass, which is at the middle of the rod. Therefore, the torque due to the weight of the rod is zero because the distance from the center of mass to each knife edge is the same.
2. The weight of the suspended mass: The 50g mass creates a torque (τ) around the supporting knife edge. We can calculate this torque using the equation τ = force × distance.
The force due to gravity is F = m × g = 0.05kg × 9.8m/s² = 0.49N,
and the distance from the suspended mass to the knife edge is 0.3m.
Therefore, the torque created by the suspended mass is τ = 0.49N × 0.3m.
Since the rod is in equilibrium, the sum of the torques on one side of the rod should be equal to the sum of the torques on the other side.
Let's denote the reaction at the first knife edge as R1 and at the second knife edge as R2.
Since the rod is of uniform mass, it is divided into two equal halves. So, the distance from each knife edge to the center of mass of the rod is 0.5m.
Using the equilibrium condition, the torque equation is:
τ1 + τ2 = 0.
Since the torque due to the weight of the rod (τ1) is zero,
τ2 = -τ = -0.49N × 0.3m.
Now, to find the reactions, we can divide the torque values by their respective distances from the knife edges.
R1 × 0.1m - R2 × 0.5m = -τ.
Plugging in the values, we get:
R1 × 0.1m - R2 × 0.5m = -(-0.49N × 0.3m).
Simplifying the equation, we find:
0.1R1 - 0.5R2 = 0.49N × 0.3m.
Finally, we have a system of two equations:
0.1R1 - 0.5R2 = 0.49N × 0.3m,
R1 + R2 = 0.49N.
Solving these equations simultaneously will give us the reactions R1 and R2.