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?
from one knife edge
center of rod is .4 meters
added mass is .3 meters
other knife edge is .8 m away from first
total mass = .1 + .05 = .15 kg
moment around close knife edge = g(.3*.05+.4*.1) = .055 g
so
F(.8) = .055 g
F = .06875 g
if g = 9.81 m/s^2
F = .674 Newtons
total weight = .15(9.81) = 1.47 N
so other F = 1.47 - .674 = .798 Newtons
Answers
To find the reaction at the supports, we need to consider the equilibrium of the rod. Since the rod is uniform and has a length of 1m, we can assume its mass is evenly distributed.
Let's label the supports as A and B, with A being the knife edge on the left and B being the knife edge on the right.
First, let's calculate the center of mass of the rod. Since the rod is uniform, the center of mass will be located at its midpoint. The distance from support A to the center of mass is 0.5m, and the distance from support B to the center of mass is also 0.5m.
Since the rod is supported horizontally, it will be in rotational equilibrium. This means that the sum of the clockwise moments about any point on the rod is equal to the sum of the anticlockwise moments about the same point.
Let's take moments about support A. The weight of the rod itself (100g) will create an anticlockwise moment, while the weight of the 50g mass (suspended 30cm from support A) will create a clockwise moment.
Clockwise moment = weight of 50g mass × distance from support A = 0.05kg × 0.3m = 0.015N·m
Anticlockwise moment = weight of the rod itself × distance from support A = 0.1kg × 0.5m = 0.05N·m
To be in equilibrium, the clockwise and anticlockwise moments must be equal:
0.015N·m = 0.05N·m
Now, let's consider the vertical forces. Since the rod is in equilibrium, the sum of the upward forces must be equal to the sum of the downward forces.
The reaction at support A will consist of two components:
1. The reaction force to balance the weight of the 50g mass, which is acting downward:
Reaction at A1 = weight of 50g mass = 0.05N
2. The reaction force to balance the weight of the rod itself, which is acting downward:
Reaction at A2 = weight of the rod itself = 0.1N
Therefore, the total reaction at support A will be the sum of these two components:
Reaction at A = Reaction at A1 + Reaction at A2 = 0.05N + 0.1N = 0.15N
Since the rod is in equilibrium, the reaction at support B will be equal to the reaction at support A:
Reaction at B = Reaction at A = 0.15N
So, the reaction at both supports A and B will be 0.15N.
To find the reaction at the supports, we need to consider the equilibrium of the rod.
Let's denote the distance between one of the knife edges and the suspended mass as "a". In this case, "a" is equal to 30cm or 0.3m.
To begin, let's calculate the torques acting on the rod.
1. Torque due to the weight of the rod:
Since the rod is uniform, we can consider its weight acting at its center of mass, which is at the midpoint of the rod.
Length of the rod (L) = 1m
Distance between center of mass and one of the knife edges (x) = (1m/2) = 0.5m
Mass of the rod (m) = 100g = 0.1kg
Gravity (g) = 9.8 m/s^2
Torque due to the weight of the rod (τ₁) = m * g * x = 0.1kg * 9.8 m/s^2 * 0.5m
2. Torque due to the weight of the suspended mass:
The suspended mass is located "a" meters from one of the knife edges.
Mass of the suspended mass (m') = 50g = 0.05kg
Distance between center of mass and the suspended mass (y) = (0.3 - 0.5)m = -0.2m (negative sign indicates that it is on the opposite side)
Torque due to the weight of the suspended mass (τ₂) = m' * g * y = 0.05kg * 9.8 m/s^2 * (-0.2m)
Next, let's consider the reaction forces at the supports.
Since the rod is in equilibrium, the sum of the torques acting on the rod must be zero. Therefore, we can write:
τ₁ + τ₂ = 0
Substituting the respective values, we get:
(0.1kg * 9.8 m/s^2 * 0.5m) + (0.05kg * 9.8 m/s^2 * (-0.2m)) = 0
Upon calculating, we find that:
∑τ = 0.49 Nm - 0.098 Nm = 0.392 Nm
Since this sum is equal to zero, it implies that the reactions at the supports are equal in magnitude and opposite in direction.
Therefore, the reaction at each support is half of the sum of the torques, which is:
Reaction at each support = (0.392 Nm) / 2 = 0.196 Nm