Question

For the given figure \( BC = 15 \, \text{cm}, BD = 35 \, \text{cm}, W = 20 \, \text{N}, W_1 = 80 \, \text{N}\), compute the net moment at the joint B. 

• A. \(19 \, \text{Nm}\)
• B. \(31 \, \text{Nm}\)
• C. \(28 \, \text{Nm}\)
• D. \(300 \, \text{Nm}\)

Hint:

In static equilibrium, the net moment is the sum of individual moments acting on a point, considering their distances from the point of rotation.

Answer 1

The Correct Option is B

To compute the moment at joint B, we use the formula for moment: \[ M = F \times d \] Where: - \(F\) is the force applied at the point, and - \(d\) is the perpendicular distance from the point of rotation (joint B). - Moment due to \(W\) at \(BC = 15 \, \text{cm} = 0.15 \, \text{m}\): \[ M_1 = W \times BC = 20 \times 0.15 = 3 \, \text{Nm} \] - Moment due to \(W_1\) at \(BD = 35 \, \text{cm} = 0.35 \, \text{m}\): \[ M_2 = W_1 \times BD = 80 \times 0.35 = 28 \, \text{Nm} \] - The net moment at joint B is the sum of the two moments: \[ M_{\text{net}} = M_1 + M_2 = 3 + 28 = 31 \, \text{Nm} \]
Conclusion: The net moment at joint B is \(31 \, \text{Nm}\), so option (b) is correct.