A ligt weightless bar is pivoted at is centre ad weight of 5n and 10n
To find the pivot point of a weightless bar with weights of 5N and 10N, we can use the principle of moments. The principle of moments states that the sum of clockwise moments is equal to the sum of anticlockwise moments for a body in equilibrium.
Since the bar is weightless, it means the weight of the bar itself can be neglected. So, we only need to consider the weights of 5N and 10N.
Let's assume the distance from the pivot point to the 5N weight is x, and the distance from the pivot point to the 10N weight is y.
To achieve equilibrium, the sum of clockwise moments must be equal to the sum of anticlockwise moments. The moment of a force is given by the product of the force and its perpendicular distance from the pivot.
So, the equation becomes:
(5N * x) = (10N * y)
To solve for the pivot point, we need to know either x or y. If we have one of the distances, we can substitute it into the equation to find the other distance.
For example, let's say the distance from the pivot to the 5N weight is 2 meters. We can substitute this into the equation:
(5N * 2m) = (10N * y)
10N = 10N * y
y = 1 meter
Therefore, the distance from the pivot point to the 10N weight is 1 meter.