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A loaded cement mixer drives onto an old drawbridge, where it stalls with its center of gravity threequarters of the way across the span. The truck driver radios for help, sets the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to the end opposite the hinge (Fig. \(\mathbf{P} 1 \mathbf{1 . 5 6}\) ). The drawbridge is \(40.0 \mathrm{~m}\) long and has a mass of \(18,000 \mathrm{~kg} ;\) its center of gravity is at its midpoint. The cement mixer, with driver, has mass \(30,000 \mathrm{~kg}\). When the drawbridge has been raised to an angle of \(30^{\circ}\) above the horizontal, the cable makes an angle of \(70^{\circ}\) with the surface of the bridge. (a) What is the tension \(T\) in the cable when the drawbridge is held in this position? (b) What are the horizontal and vertical components of the force the hinge exerts on the span?

Step by step solution

01

Calculate Torque Due to Tension

Find the distance from the hinge to where the cable is attached (which is the length of the bridge as the cable is attached to the end of the bridge opposite the hinge). Calculate the force due to tension in the bridge using the formula to calculate torque which is the product of the force and the distance from the pivot point (or hinge). Torque due to tension \(TT\) can be written as follows: \( TT = T \cdot L \cdot cos(70) \) where \(L = 40m\) is the length of the bridge, \(T\) is the tension in the cable, and \(cos(70)\) comes from the angle between the tension force and the horizontal.

02

Calculate Torque Due to Weight of the Bridge

Calculate the force due to the weight of the bridge also using the torque formula. This can be written as \( TW = W_b \cdot (L/2) \cdot sin(30) \), where \(W_b = 18000kg \cdot g\) is the weight of the bridge, \((L/2)\) is the distance from the hinge to the center of gravity of the bridge, and \(sin(30)\) comes from the angle between the weight of the bridge and the horizontal.

03

Calculate Torque Due to Weight of the Truck

Finally, calculate the force due to the weight of the truck again using the torque formula. This can be written as \( TTu = W_T \cdot (3L/4) \cdot sin(30) \), where \(W_T = 30000kg \cdot g\) is the weight of the truck, \((3L/4)\) is the distance from the hinge to the center of gravity of the truck, and \(sin(30)\) comes from the angle between the weight of the truck and the horizontal.

04

Calculate the Tension

As the drawbridge is in static equilibrium, the sum of all the torques about the hinge must be equal to zero. So, equating \( TT = TW + TTu \) and solving for \( T \).

05

Calculate Horizontal and Vertical Components

The force \( F \) the hinge exerts on the span can be divided into vertical (\( F_y \)) and horizontal (\( F_x \)) components, found by using Newton's 2nd law. Vertically, this gives \( F_y = W_T + W_B - T \cdot sin(70) \). Horizontally, this gives \( F_x = T \cdot cos(70) \). Solve these to get the horizontal and vertical components.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Equilibrium

Static equilibrium occurs when all the forces and torques acting on a body are balanced, meaning the object is at rest or moving at a constant velocity. In the scenario of the drawbridge and the cement mixer, the drawbridge remains steady due to this balance. Static equilibrium ensures that the sum of all torques and forces acting on the bridge must equal zero. This is why the concept is pivotal when determining the tension in the cable and the forces acting at the hinge.
For the drawbridge:

• The gravitational forces on the bridge and the cement mixer create torques that want to rotate the bridge around the hinge.
• The tension in the cable acts to keep the bridge from rotating, creating an opposing torque.

To solve the problem, we need to sum up the torques produced by these forces and set them equal to zero. This is the condition of static equilibrium that allows us to calculate the tension in the cable.

Tension

Tension refers to the force exerted along a stretched cable or rope. In the context of the drawbridge scenario, the tension in the cable keeps the bridge raised. Calculating this tension involves understanding how the forces translate to torques around the hinge point of the bridge.
The process involves:

• Considering the angle of the cable with the bridge (70 degrees) to find the horizontal component of the tension since the tension force isn't acting purely horizontally or vertically.
• Using the torque equation which incorporates tension (\(TT = T \cdot L \cdot \cos(70)\)), where \(L\) is the total length of the bridge since the cable is attached at the bridge’s end.
• Balancing this torque against the torques due to the weights of the drawbridge and the cement mixer to keep the system in equilibrium.

Solving this equation gives us the tension in the cable needed to hold the bridge at the given angle.

Newton's Second Law

Newton's Second Law states that the acceleration of an object is dependent on the net forces acting upon it and the object’s mass. For the drawbridge in question, there is no acceleration—it’s in static equilibrium. We use Newton’s Second Law to ensure all forces balance out without causing acceleration.
In terms of forces on the drawbridge:

• The weight of the bridge and cement mixer exert downward forces.
• The hinge exerts supporting forces to counteract these weights as well as the force by the tension in the cable, split into horizontal and vertical components.

Using the law,
Vertical forces balance when considering gravitational forces and tension’s vertical component:\(F_y = W_T + W_B - T \cdot \sin(70)\)
Horizontal forces involve only the cable tension:\(F_x = T \cdot \cos(70)\)
By solving these equations, you can find out how much force the hinge exerts to keep the bridge from moving.

Center of Gravity

The center of gravity is crucial in analyzing static equilibrium problems as it’s the point where gravitational force can be assumed to act. For the drawbridge and cement mixer:

• The bridge's center of gravity is at its midpoint, which simplifies calculations as forces and torques can be considered as acting from this point.
• The cement mixer is positioned three-quarters across the bridge, moving its center of gravity further from the hinge.

Understanding these positions helps in calculating the torque each weight induces. The farther a force acts from the hinge, the more it influences the torque:

• For the bridge, torque is calculated using half its length (\(L/2\)) due to its central center of gravity.
• The cement mixer adds additional torque calculated using \(3L/4\), since it is three-quarters down the bridge, adding significant rotational force around the hinge.

This understanding is key in setting up the equilibrium equations needed to solve for the tension and forces in the problem.