A light bar is pivoted at it's centre and weight of 5n and 10n, 3m and 2m from the pivot on one side ate balanced by a weight of 20n on the other side. How far is the 20n weight from the pivot
5*3 + 10*2 = 20x
To find the distance of the 20N weight from the pivot point, we can use the principle of moments.
The principle of moments states that for an object in rotational equilibrium, the sum of the anticlockwise moments about a point is equal to the sum of the clockwise moments about the same point.
In this case, the moments on each side of the pivot need to balance each other out.
On one side of the pivot, we have a weight of 5N at a distance of 3m, and a weight of 10N at a distance of 2m. The total moment on this side would be: 5N * 3m + 10N * 2m.
On the other side of the pivot, we have a weight of 20N which we need to find the distance for (let's call it x).
So, the total moment on the other side would be: 20N * x.
According to the principle of moments, these two total moments should be equal:
5N * 3m + 10N * 2m = 20N * x.
Let's solve for x:
15N + 20N = 20N * x.
Simplifying the equation:
35N = 20N * x.
Dividing both sides by 20N:
35N / 20N = x.
1.75 = x.
Therefore, the 20N weight is located 1.75 meters away from the pivot point.