A uniform rod,one metre long of mass 50g is supported horizontally on 2 knife edges placed 10cm from it's ends. what will be the direction at this Support when a hundred G mass is suspended 10cm from the mid-point of the rod(in Newton)
In order to determine the direction of the support when a 100g mass is suspended, we need to consider the torques acting on the rod.
First, calculate the moment of inertia of the rod about one of its ends using the parallel axis theorem. Since the rod is uniform, the moment of inertia about an end can be given as:
I = (1/3) * m * L^2
where:
I is the moment of inertia
m is the mass of the rod (50g = 0.05kg)
L is the length of the rod (1m)
I = (1/3) * 0.05kg * (1m)^2
I = 0.05/3 kgm^2
Next, calculate the clockwise torque (τc) due to the weight of the rod about the left support. The weight of the rod acts at its center of mass, which is at the midpoint of the rod. The distance between the left support and the center of mass is 0.1m.
τc = m * g * d
where:
τc is the torque
m is the mass of the rod (0.05kg)
g is the acceleration due to gravity (9.8m/s^2)
d is the distance between the left support and the midpoint (0.1m)
τc = 0.05kg * 9.8m/s^2 * 0.1m
τc = 0.049 Nm
Now, consider the anticlockwise torque (τac) due to the 100g mass. The mass is located 0.1m from the midpoint of the rod.
τac = m * g * d
where:
τac is the torque
m is the mass of the 100g weight (0.1kg)
g is the acceleration due to gravity (9.8m/s^2)
d is the distance between the midpoint and the mass (0.1m)
τac = 0.1kg * 9.8m/s^2 * 0.1m
τac = 0.098 Nm
Since the rod is in equilibrium, the sum of the torques acting on it must be zero. Therefore, the direction of the support will be determined by the torque acting on it.
Sum of torques = τac - τc
Sum of torques = 0.098 Nm - 0.049 Nm
Sum of torques = 0.049 Nm
The direction of the support will be in the clockwise direction, as the sum of torques is positive.