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Chapter 9: Problem 82

A long, uniform rope with a mass of 0.135 kg per meter lies on the ground. You grab one end of the rope and lift it at the constant rate of \(1.13 \mathrm{m} / \mathrm{s}\). Calculate the upward force you must exert at the moment when the top end of the rope is \(0.525 \mathrm{m}\) above the ground.

Short Answer

Expert verified
The upward force is approximately 0.695 newtons.

Step by step solution

01

Calculate the Mass of the Lifted Rope

First, determine the mass of the portion of the rope being lifted. The top end of the rope is 0.525 meters above the ground. Given the rope's mass per unit length is 0.135 kg/m, we calculate the mass of the lifted portion as \( m = 0.135 \times 0.525 \).
02

Calculate the Weight of the Lifted Rope

Next, calculate the gravitational force acting on the lifted rope, which is its weight. This is calculated using the formula \( F_G = m \cdot g \), where \( g = 9.81 \; \text{m/s}^2 \) is the acceleration due to gravity. Substitute the mass from Step 1 and calculate \( F_G = 0.070875 \times 9.81 \).
03

Calculate the Required Upward Force

Since you are lifting the rope at a constant speed, the net force acting on the rope is zero, i.e., the upward force you exert must equal the gravitational force downwards. Thus, the upward force \( F_{up} \) is equal to the weight calculated in Step 2.

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Key Concepts

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

Mass Calculation
In physics, determining the mass of an object is essential for many calculations. Here, we consider a rope with a mass per length of 0.135 kg/m. To calculate the mass of the portion of the rope you are lifting off the ground, multiply the given mass per meter by the length of the rope that is elevated. When you know the rope is elevated to 0.525 meters, the calculation becomes simple:
  • Mass of lifted rope = 0.135 kg/m
  • Length of rope above ground = 0.525 m
Therefore, the total mass is given by: \( m = 0.135 \times 0.525 = 0.070875 \; \text{kg} \).
Understanding the mass is fundamental for subsequent steps in physics calculations involving forces.
Gravitational Force
Gravitational force is the force that the Earth exerts to pull objects towards its center. Every object with mass experiences gravitational force. For any object near the Earth's surface, this force can be calculated using the formula \( F_G = m \cdot g \),where
  • \( m \) is the mass of the object, and
  • \( g \approx 9.81 \; \text{m/s}^2 \)
this is the acceleration due to gravity.
Given the previously calculated mass of the lifted rope (0.070875 kg), the gravitational force acting on it is:
\( F_G = 0.070875 \times 9.81 = 0.69527375 \; \text{N} \).
This force represents the weight of the lifted rope and is crucial for calculating any action to counterbalance it.
Constant Speed
When lifting an object at constant speed, the notion of forces becomes significant. Constant speed indicates that the object does not accelerate or decelerate, implying that the applied force counteracts other forces precisely.
  • In mechanics, this means the net force is zero.

For the given exercise, this principle is crucial. The upward force you apply must equal the downward gravitational force on the rope. Any variation in speed, be it accelerating up or slowing down, would influence the necessary force, due to adjustments in kinetic energy or work done.
Net Force
Net force is the total force acting on an object and determines its motion. A net force of zero implies that the object maintains a uniform state 鈥� either staying still or moving at constant velocity. In the context of lifting the rope at constant speed:
  • Upward force = Downward gravitational force
  • Net force = 0
When the forces are balanced, as in this scenario, you achieve constant speed. Remember, any net force not equal to zero would cause an object to either accelerate or decelerate. Therefore, understanding net force is essential when calculating how to maintain or alter the motion of objects in mechanics.
Upward Force
Exerting an upward force supports lifting objects against gravity. To keep the rope moving upwards at a constant rate, the force you exert must perfectly balance the gravitational force pulling it back downwards. This principle is central to scenarios where maintaining a steady state of motion is necessary.
  • To find the upward force \( F_{up} \) to apply, recognize it must equal the rope鈥檚 weight.
  • For the lifted rope: \( F_{up} = 0.69527375 \; \text{N} \)

The equal and opposite forces result in a net force of zero, ensuring the constant velocity of the lift. For practical applications, ensure the upward force is sustained precisely to match the changing gravitational force as conditions vary.

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Most popular questions from this chapter

A 26.2-kg dog is running northward at 2.70 m/s, while a \(5.30-\mathrm{kg}\) cat is running eastward at \(3.04 \mathrm{m} / \mathrm{s} .\) Their \(74.0-\mathrm{kg}\) owner has the same momentum as the two pets taken together. Find the direction and magnitude of the owner's velocity.

A hoop of mass \(M\) and radius \(R\) rests on a smooth, level surface. The inside of the hoop has ridges on either side, so that it forms a track on which a ball can roll, as indicated in Figure \(9-25\). If a ball of mass \(2 M\) and radius \(r=R / 4\) is released as shown, the system rocks back and forth until it comes to rest with the ball at the bottom of the hoop. When the ball comes to rest, what is the \(x\) coordinate of its center?

A 0.14-kg baseball moves toward home plate with a velocity \(\overrightarrow{\mathbf{v}}_{i}=(-36 \mathrm{m} / \mathrm{s}) \hat{\mathrm{x}}\). After striking the bat, the ball moves vertically upward with a velocity \(\overrightarrow{\mathbf{v}}_{\mathrm{f}}=(18 \mathrm{m} / \mathrm{s}) \hat{\mathrm{y}} .\) (a) Find the direction and magnitude of the impulse delivered to the ball by the bat. Assume that the ball and bat are in contact for \(1.5 \mathrm{ms}\). (b) How would your answer to part (a) change if the mass of the ball were doubled? (c) How would your answer to part (a) change if the mass of the bat were doubled instead?

A bullet with a mass of 4.0 g and a speed of \(650 \mathrm{m} / \mathrm{s}\) is fired at a block of wood with a mass of \(0.095 \mathrm{kg}\). The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is \(23 \mathrm{m} / \mathrm{s}\). (a) What is the speed of the bullet when it exits the block? (b) Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy? Explain. (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

An object initially at rest breaks into two pieces as the result of an explosion. One piece has twice the kinetic energy of the other piece. What is the ratio of the masses of the two pieces? Which piece has the larger mass?
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