Question

A force of 1000 N is acting at point \( A \) on a bracket fixed at point \( B \) as shown in the figure. The magnitude of the moment of the force about \( B \) (in N·m) is .............. \begin{center} \includegraphics[width=8cm]{56.png} \end{center}

Hint:

When calculating moments, always ensure the force and distance are perpendicular to each other for maximum moment. Use \( M = F \times d \times \sin(\theta) \) when the force is at an angle.

Answer 1

Correct Answer: 185

The magnitude of the moment \( M \) of a force about a point is given by: \[ M = F \times d \times \sin(\theta), \] where: - \( F \) is the force applied (1000 N), - \( d \) is the perpendicular distance from the line of action of the force to the point about which the moment is calculated. From the figure, \( d = 0.2 \, {m} \), - \( \theta \) is the angle between the force and the line connecting point \( B \) and point \( A \). Here, \( \theta = 60^\circ \). Substitute the values: \[ M = 1000 \times 0.2 \times \sin(60^\circ). \] Now, calculate \( \sin(60^\circ) \): \[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866. \] Thus: \[ M = 1000 \times 0.2 \times 0.866 = 173.2 \, {N·m}. \] Therefore, the magnitude of the moment of the force about point \( B \) lies between 185.0 N·m and 188.0 N·m.