A uniform meter rod of weight 40N has a piece of metal of weight 20N attached to one of its end. The centre of gravity of system from the weighted end is?
To find the center of gravity of the system, we need to consider the weight and the position of the individual components.
Given:
Weight of the meter rod = 40N
Weight of the attached metal piece = 20N
Since weight is a force, it acts at a specific point, which is the center of gravity. As the meter rod is uniform in shape, the center of gravity of the rod will be at its midpoint.
The center of gravity of the weighted end can be found using the following formula:
(center of gravity of system) = (total moment of weight 1 + total moment of weight 2) / (total weight)
The total weight is the sum of the weight of the meter rod and the weight of the attached metal piece: 40N + 20N = 60N.
The total moment of weight 1 (40N) will be zero because the center of gravity of the rod is at its midpoint.
The total moment of weight 2 (20N) will be calculated based on the position of the weighted end from the midpoint. Since the weight is attached to one end, the distance from the weighted end to the midpoint is half of the length of the meter rod, which is 1 meter/2 = 0.5 meters.
Therefore, the total moment of weight 2 = weight of weight 2 x distance to midpoint = 20N x 0.5m = 10Nm.
Using the formula mentioned earlier:
(center of gravity of system) = (0 + 10Nm) / 60N = 10Nm / 60N = 0.167m.
Hence, the center of gravity of the system from the weighted end is 0.167 meters.
To find the center of gravity of the system, we need to consider the weights of the meter rod and the attached metal piece.
Given:
Weight of the meter rod = 40N
Weight of the attached metal piece = 20N
Since the metal piece is attached to one end of the meter rod, the distance between the center of gravity of the meter rod and the weighted end will be the length of the meter rod.
Since a meter rod is a uniform rod, we assume that the weight is evenly distributed along its length. Therefore, the center of gravity of the meter rod will be at its midpoint.
So, the center of gravity of the meter rod will be at a distance of 1 meter/2 = 0.5 meters from the weighted end.
Therefore, the center of gravity of the system from the weighted end is 0.5 meters.
center of mass of rod at 0.50 m
take moments about weighted end
20 * 0 + 40 * 0.5 = 60 x
x = (2/3) meter