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A \(75.0 \mathrm{~kg}\) wrecking ball hangs from a uniform, heavy-duty chain of mass \(26.0 \mathrm{~kg}\). (a) Find the maximum and minimum tensions in the chain. (b) What is the tension at a point three-fourths of the way up from the bottom of the chain?

Short Answer

Expert verified
The maximum tension in the chain is \(990.2N\), the minimum tension is \(735N\), and the tension at a point three-fourths of the way up from the bottom of the chain is \(798.7N\).

Step by step solution

01

Max and Min Chain Tensions

The maximum tension in the chain is at the topmost link, which bears the weight of both the wrecking ball and the chain. The weight of an object is given by \( mg \), where \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity (9.8m/s^2). Adding the weight of the wrecking ball \( (75.0 kg * 9.8 m/s^2) \) and the chain \( (26.0 kg * 9.8 m/s^2) \) gives \( 990.2 N \). The minimum tension in the chain is at the bottommost link, which bears only the weight of the wrecking ball, hence, \( 75.0 kg * 9.8 m/s^2 \) = \( 735.0 N \).
02

Tension Three-Fourths Up the Chain

The tension at a point three-fourths of the way up the chain is due to the weight of the wrecking ball and the weight of the lower one-fourth of the chain (since the tension at any point in the chain is due to the weight below that point). The weight of one-fourth of the chain is \( 0.25 * 26.0 kg * 9.8 m/s^2 = 63.7 N \). Therefore, the tension at this point is \( 735.0 N + 63.7 N \) = \( 798.7 N \).

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Key Concepts

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

Tension in a Chain
Understanding the concept of tension in a chain is crucial when trying to solve physics problems related to it. Tension refers to the force transmitted through a string, cable, or in this case, a chain, when it is pulled tight by forces acting from opposite ends. In practical terms, think of tension as a kind of 'stretching' force that is trying to pull the chain apart. The actual tension at any point in the chain depends on the forces applied to it and the mass of the objects hanging from it.

In the example of the wrecking ball hanging from a chain, the top of the chain experiences the highest tension because it must support the weight of the entire wrecking ball and the rest of the chain below it. As we move down the chain link by link, the tension decreases incrementally because each link below supports less weight. This concept is central to many practical applications involving chains, cables, or ropes, whether in construction, engineering, or understanding basic physical situations.
Weight Calculation
The weight of an object is a fundamental concept in physics, and understanding how to calculate it is essential for solving problems involving forces and tension. Weight is the force exerted by gravity on an object. It is mathematically determined by the equation \( W = m \times g \), where \( W \) is the weight, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately \( 9.8 \, m/s^2 \).

When dealing with chains and weights, as in the wrecking ball scenario, the weight of the chain must be calculated and added to the weight of the attached object to find the maximum tension at the top. A crucial consideration is that the weight is not the same as mass, although they are directly proportional; weight is a force while mass is a measure of an object's inertia.
Uniform Chain Physics
A uniform chain is one where the mass is evenly distributed throughout its length. In physics problems, this assumption allows us to simplify calculations because every segment of the chain can be treated as having the same mass per unit length. This is especially useful when we need to find the tension at different points in the chain.

When a chain is uniform, the tension at a point in the chain is affected only by the weight of the chain hanging below that point and any additional weights it may be supporting. For example, if we have to calculate the tension three-fourths of the way up a hanging uniform chain, we only consider the weight of the remaining one-fourth of the chain below that point and the weight of any objects it is supporting.
Acceleration Due to Gravity
The acceleration due to gravity, often denoted as \( g \), is a physical constant that measures the rate of acceleration of objects as they fall freely towards a massive body like Earth. On Earth, this value is approximately \( 9.8 \, m/s^2 \). This acceleration is the same for all objects, regardless of their mass, provided that air resistance can be neglected.

When we calculate the weight of the wrecking ball or the chain in our example, we use the formula \( W = m \times g \) to include the effect of gravity. This factor is crucial in physics problems involving forces and motion because it influences how objects behave under the influence of Earth's gravity. Consideration of \( g \) allows us to predict the behavior of objects in free fall, calculate the forces in objects at rest like our hanging chain, and understand the principles that govern motion on our planet.

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Most popular questions from this chapter

A racetrack curve has radius \(90.0 \mathrm{~m}\) and is banked at an angle of \(18.0^{\circ} .\) The coefficient of static friction between the tires and the roadway is \(0.400 .\) A race car with mass \(1200 \mathrm{~kg}\) rounds the curve with the maximum speed to avoid skidding. (a) As the car rounds the curve, what is the normal force exerted on it by the road? What are the car's (b) radial acceleration and (c) speed?

DATA A road heading due east passes over a small hill. You drive a car of mass \(m\) at constant speed \(v\) over the top of the hill, where the shape of the roadway is well approximated as an arc of a circle with radius \(R\). Sensors have been placed on the road surface there to measure the downward force that cars exert on the surface at various speeds. The table gives values of this force versus speed for your car: $$ \begin{array}{lcccccc} \text { Speed }(m / s) & 6.00 & 8.00 & 10.0 & 12.0 & 14.0 & 16.0 \\ \hline \text { Force }(\mathrm{N}) & 8100 & 7690 & 7050 & 6100 & 5200 & 4200 \end{array} $$ Treat the car as a particle. (a) Plot the values in such a way that they are well fitted by a straight line. You might need to raise the speed, the force, or both to some power. (b) Use your graph from part (a) to calculate \(m\) and \(R .\) (c) What maximum speed can the car have at the top of the hill and still not lose contact with the road?

Friction in an Elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{~m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of \(36.0 \mathrm{~kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level bed of the truck are 0.355 and 0.650 , respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to \(30.0 \mathrm{~m} / \mathrm{s}\) without causing the box to slide? Draw a free-body diagram of the toolbox.

A man pushes on a piano with mass \(180 \mathrm{~kg} ;\) it slides at constant velocity down a ramp that is inclined at \(19.0^{\circ}\) above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor.

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