A uniform beam AB 4cm long is pivoted at its geometrical centre and is held in equilibrium by a number of weights suspended as shown below.Determine the value of distance X.

Answer 1

To find the value of distance X, we need to consider the moments acting on the beam about the pivot point.

Let the weight on the left side of the pivot be W1 and the weight on the right side be W2. The distances of these weights from the pivot are 2 cm each.

Since the beam is in equilibrium, the sum of the anti-clockwise moments about the pivot must equal the sum of the clockwise moments about the pivot.

W1 * 2 = W2 * 2

Given that W1 = 5N and W2 = 10N, we can now solve for the value of distance X:

5 * 2 = 10 * X

10 = 10X

X = 1

Therefore, the value of distance X is 1 cm.