91Ó°ÊÓ

When considering the scenario where the drawbridge cable suddenly breaks, it is crucial to understand the concept of angular acceleration. Angular acceleration, represented by the symbol \( \alpha \), is the rate at which the angular velocity of an object changes with time. In this problem, once the cable breaks, the drawbridge will start to rotate about the hinge due to the unbalanced gravitational force acting on its center of gravity.

For a rod-like object such as a drawbridge rotating about one end, the moment of inertia \( I \) is given by the formula:

• \( I = \frac{1}{3} m L^2 \)

where \( m \) is the mass of the drawbridge and \( L \) is its length.
If we then apply Newton's second law for rotation (Torque = Moment of Inertia \( \times \) Angular Acceleration), we can express the angular acceleration \( \alpha \) as:

• \( \tau = I \alpha \)

The torque \( \tau \) caused by the weight of the drawbridge when the cable breaks is calculated as the force of gravity acting at the center of gravity multiplied by the perpendicular distance from the pivot, which is:

• \( \tau = 45000 \times 7 \sin(37^{\circ}) \)

By substituting this into the equation for torque, the angular acceleration \( \alpha \) can be determined.

Static equilibrium is a state where a rigid body remains at rest, and all forces and torques acting on it are balanced. In the drawbridge problem, the bridge is held in static equilibrium when it is at a \(37^{\circ}\) angle. This means the sum of all forces and the sum of all torques acting on the bridge must be zero. This is achieved by balancing the torque caused by the weight of the drawbridge with the tension from the cable exerting force along the bridge.

To determine equilibrium conditions, consider:

• The sum of horizontal forces: Tension forces must equal the horizontal component of hinge forces.
• The sum of vertical forces: The vertical component of hinge forces minus the component of weight acting vertically should sum to zero.
• The sum of torques about the hinge must be zero: The torque from the cable tension must counterbalance the torque due to the weight of the bridge.

Mathematically, this is expressed as:

• \( T \times 3.5 \cos(37^{\circ}) = 45000 \times 7 \sin(37^{\circ}) \)

This ensures the bridge remains in its designated position until external conditions change.

Understanding how forces interact at hinges is key in statics, especially in scenarios involving rotating bodies such as drawbridges. Hinge force components consist of horizontal and vertical forces exerted by the hinge that keep the structure stabilized. These forces counteract the external forces applied, such as tension from a cable or weight due to gravity.

To calculate these components in the drawbridge problem, the following steps are taken:

• **Horizontal Component** \( (F_{hx}) \): Equal to the tension in the cable as it acts horizontally.
• **Vertical Component** \( (F_{hy}) \): Equal to the total weight of the drawbridge minus the force component due to tension:

\[F_{hy} = W - T \sin(37^{\circ})\]Together, these components will give the net force and direction exerted by the hinge to maintain the drawbridge's static position. Being able to decompose these forces offers insights into how designs can withstand loads and maintain structural integrity.