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Chapter 5: Problem 1

Two 25.0 -N weights are suspended at opposite ends of a rope that passes over a light, frictionless pulley. The pulley is attached to a chain that goes to the ceiling. (a) What is the tension in the rope? (b) What is the tension in the chain?

Short Answer

Expert verified
(a) 25.0 N; (b) 50.0 N.

Step by step solution

01

Understanding the System

In this problem, we have a system where two 25.0 N weights are hanging on opposite sides of a rope over a pulley. The system is symmetrical 鈥� each weight pulls down on the rope with a force equal to their weight.
02

Analyzing Forces on the Rope

The system is in equilibrium because the weights are equal and opposite. The tension in any rope section is equal to the force exerted by each weight. Since both weights are equal to 25.0 N, the tension in the rope is also 25.0 N.
03

Calculating Tension in the Rope

Since the tension throughout the rope must be the same, and both weights are pulling equally, the tension in the rope is directly equal to the weight of either weight. Thus, the tension in the rope is 25.0 N.
04

Understanding Tension in the Chain

The chain supports the pulley and the weights. For the system to be in equilibrium, the chain must provide enough force to balance the total downward force exerted by the weights. This means it must compensate for both weights.
05

Calculating Total Force on the Chain

The total tension in the chain is the sum of the forces from both weights, which is 25.0 N from each weight. Therefore, the total force exerted on the chain from the weights is 50.0 N.

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Key Concepts

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

Equilibrium in Physics
When we talk about equilibrium in physics, we're referring to a state where all forces acting on a system are perfectly balanced. In simpler terms, equilibrium is the balance point that keeps a system steady and unmoving. In our pulley system problem, the key is that each 25.0 N weight pulls equally on the rope, creating a state of equilibrium. For equilibrium to be achieved, the forces must be equal in size but opposite in direction, ensuring the system doesn't unintentionally accelerate or move. Without this balance, the weights in the pulley system would tip in one direction or the other, disrupting equilibrium and causing motion.
Force Balance
Force balance is crucial in understanding how systems like our pulley problem stay still. Imagine pulling on either side of a rope with equal but opposite forces. These forces cancel each other out, maintaining a stationary position. This is exactly what's happening in our exercise. The 25.0 N weights each pull down on the rope, yet because these forces are equal and opposing, the rope doesn't move.
  • The force from the first weight is 25.0 N downward.
  • The force from the second weight is also 25.0 N downward.
Because these forces are balanced, the tension felt throughout the rope is the same at every point. This force balance concept explains why the rope tension matches the force from a single weight: 25.0 N.
Pulley Systems
Pulley systems, like in our exercise, offer a fantastic way to understand how force is distributed and balanced. A pulley consists of a wheel around which a rope can move, changing the direction of the force exerted on it. In our problem, the pulley is frictionless and light, which means it doesn't affect the tension calculations directly. Instead, it simply changes the direction of the forces acting on the weights.

When running through these kinds of problems:
  • Recognize that a single rope in the system means the tension is the same throughout.
  • The chain attached to the ceiling supports the combined weight, thus the tension in the chain adds up to 50.0 N, balancing both hanging weights.
  • In real-world applications, understanding these mechanics helps in designing efficient systems for lifting or holding weights with minimal effort.
Pulley systems effectively distribute forces, minimizing the required output to achieve balance.

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Most popular questions from this chapter

While a person is walking, his arms swing through approximately a \(45^{\circ}\) angle in \(\frac{1}{2}\) s. As a reasonable approximation, we can assume that the arm moves with constant speed during each swing. A typical arm is 70.0 \(\mathrm{cm}\) long, measured from the shoulder joint. (a) What is the acceleration of a \(1.0-\mathrm{g}\) drop of blood in the fingertips at the bottom of the swing? (b) Draw a free-body diagram of the drop of blood in part (a). ( c)Find the force that the blood vessel must exert on the drop of blood in part (a). Which way does this force point? (d) What force would the blood vessel exert if the arm were not swinging?

(a) If the coefficient of kinetic friction between tires and dry pavement is \(0.80,\) what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

You are driving a classic 1954 Nash Ambassador with a friend who is sitting to your right on the passenger side of the front seat. The Ambassador hat bench seats. You would like to be closer to your friend and decide to use physics to achieve your romantic goal by making a quick turn. (a) Which way (to the left or to the right) should you turn the car to get your friend to slide closer to you? (b) If the coefficient of static friction between your friend and the car seat is \(0.35,\) and you keep driving at a constant speed of \(20 \mathrm{m} / \mathrm{s},\) what is the maximum radius you could make your turn and still have your friend slide your way?

Block \(A,\) with weight \(3 w,\) slides down an inclined plane \(S\) of slope angle \(36.9^{\circ}\) at a constant speed while plank \(B\) with weight \(w,\) rests on top of A. The plank is attached by a cord to the wall (Fig. P5.99). (a) Draw a diagram of all the forces acting on block A. (b) If the coefficient of kinetic friction is the same between \(A\) and \(B\) and between \(S\) and \(A,\) determine its value.

You are riding your motorcycle one day down a wet street that slopes downward at an angle of \(20^{\circ}\) below the horizontal. As you start to ride down the hill, you notice a construction crew has dug a deep hole in the street at the bottom of the hill. A siberian tiger, escaped from the City Zoo, has taken up residence in the hole. You apply the brakes and lock your wheels at the top of the hill, where you are moving with a speed of 20 \(\mathrm{m} / \mathrm{s} .\) The inclined street in front of you is 40 \(\mathrm{m}\) long. (a) Will you plunge into the hole and become the tiger's lunch, or do you skid to a stop before you reach the hole? (The coefficients of friction between your motorcycle tires and the wet pavement are \(\mu_{\mathrm{s}}=0.90\) and \(\mu_{\mathrm{k}}=0.70 .\) ) ( b ) What must your initial speed be if you are to stop just before reaching the hole?
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