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Chapter 5: Problem 19

Atwood's Machine. A \(15.0-\mathrm{kg}\) load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0 \(\mathrm{kg}\) counterweight is suspended from the other end of the rope as shown in Fig. 5.51 The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the tensionin the rope while the load is moving? How does the tension compare to the weight of the load of bricks? To the weight of the counterweight? figure can't copy

Short Answer

Expert verified
The bricks accelerate at 4.08 m/s虏 upward, with a rope tension of 208.2 N.

Step by step solution

01

Analyze the Forces on the Bricks

Consider the load of bricks with mass \( m_1 = 15.0 \text{ kg} \). The forces acting on the bricks are the gravitational force \( F_{g_1} = m_1 g \) (acting downward) and the tension \( T \) (acting upward). The equation for the net force on the bricks is: \[ m_1 a = T - m_1 g \].
02

Analyze the Forces on the Counterweight

Consider the counterweight with mass \( m_2 = 28.0 \text{ kg} \). The forces acting on the counterweight are the gravitational force \( F_{g_2} = m_2 g \) (acting downward) and the tension \( T \) (acting upward). The equation for the net force on the counterweight is: \[ m_2 a = m_2 g - T \].
03

Calculate the Acceleration

Add the two equations from Steps 1 and 2. This eliminates \( T \) and allows for solving for \( a \):\[ m_1 a + m_2 a = m_2 g - m_1 g \]\[ (m_1 + m_2) a = (m_2 - m_1) g \]\[ a = \frac{(m_2 - m_1) g}{m_1 + m_2} \]Plug in the known values: \( g = 9.8 \text{ m/s}^2 \), \( m_1 = 15.0 \text{ kg} \), \( m_2 = 28.0 \text{ kg} \).\[ a = \frac{(28.0 - 15.0) \times 9.8}{28.0 + 15.0} = 4.08 \text{ m/s}^2 \].
04

Calculate the Tension

Use the equation from Step 1 to find the tension \( T \):\[ T = m_1 g + m_1 a \]Plug in \( m_1 = 15.0 \text{ kg} \), \( a = 4.08 \text{ m/s}^2 \), and \( g = 9.8 \text{ m/s}^2 \):\[ T = 15.0 \times 9.8 + 15.0 \times 4.08 = 208.2 \text{ N} \].

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Key Concepts

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

Free-Body Diagrams
A free-body diagram is a graphical illustration used to visualize the forces acting on an object. In Atwood's Machine, each mass (bricks and counterweight) has its own free-body diagram. For the load of bricks, we identify two forces: the gravitational force
  • The downward force due to gravity: \( F_{g_1} = m_1 g \)
  • The upward force of tension in the rope: \( T \)
Similarly, for the counterweight, we consider:
  • The downward force due to gravity: \( F_{g_2} = m_2 g \)
  • The upward force of tension: \( T \)
The free-body diagrams visually represent these forces and are crucial for setting up the equations of motion in problems like Atwood's Machine. Drawing these diagrams helps in understanding how the two masses interact through the tension in the rope and how their weights lead to acceleration.
Acceleration
The acceleration in an Atwood's Machine is what causes the movement of both mass systems. To find it, we first need to understand that it is due to the net force acting on the system, which is the difference in the gravitational forces on the two masses. When both objects move, one will accelerate upwards, and the other will naturally accelerate downwards.Using Newton's second law, we combine the forces on both masses to solve for the acceleration \( a \). The key formula used is:\[a = \frac{(m_2 - m_1)g}{m_1 + m_2}\]The above formula shows that the acceleration depends on the gravitational pull (\(g\)) and the difference in mass between the two objects, as well as their total mass.
  • If both masses were equal, \( m_1 = m_2 \), the acceleration \( a \) would be zero because the forces are balanced.
  • For any difference in mass, \(a\) will have a non-zero value, moving the system.
This acceleration leads to the motion of the entire system, causing one mass to rise and the other to fall.
Tension in Rope
In a system like Atwood's Machine, the tension in the rope plays a vital role in connecting the two masses and transmitting force. The tension is essentially the force trying to "hold" the objects together in equilibrium.From the analysis of each mass, we use equations to solve for the tension \( T \). For the bricks, we rearrange the formula to define tension in terms of the given variables and calculated acceleration:\[ T = m_1 g + m_1 a\]This equation states that tension equals the sum of the forces due to gravity and the backward force from acceleration. In practice:
  • If the system is accelerating upwards, tension is greater than the gravitational force on the lighter mass.
  • Conversely, the tension would be less than gravitational force on the heavier mass.
By understanding the tension in the rope, you can see how it affects the individual component鈥檚 motion and balance within the system.

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Most popular questions from this chapter

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?

A box with mass \(m\) is dragged across a level fioor having a coefficient of kinctic friction \(\mu_{k}\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F .(\text { a) In }\) terms of \(m, \mu_{k}, \theta,\) and \(g,\) obtain an expression for the magnitude of force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a \(90-\mathrm{kg}\) patient across a floor at constant speed by pulling on him at an angle of \(25^{\circ}\) above the borizontal. By dragging some weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_{k}=0.35 .\) Use the result of part (a) to answer the instructor's question.

A \(25.0-\mathrm{kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .\) (a) As the angle \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.0 \(\mathrm{m}\) along the loading ramp?

Uterior Motives. 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 has flat 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?

A model airplane with mass 2.20 \(\mathrm{kg}\) moves in the \(x y\) -plane such that its \(x-\) and \(y\) -coordinates vary in time according to \(x(t)=\alpha-\beta t^{3}\) and \(y(t)=y t-\delta t^{2},\) where \(\alpha=1.50 \mathrm{m}, \beta=\) \(0.120 \mathrm{m} / \mathrm{s}^{3}, \gamma=3.00 \mathrm{m} / \mathrm{s},\) and \(\delta=1.00 \mathrm{m} / \mathrm{s}^{2} .\) (a) Calculate the \(x-\) and \(y\) -components of the net force on the plane as functions of time. (b) Sketch the trajectory of the airplane between \(t=0\) and \(t=3.00 \mathrm{s}\) , and draw on your sketch vectors showing the net force on the airplane at \(t=0, t=1.00 \mathrm{s}, t=2.00 \mathrm{s},\) and \(t=3.00 \mathrm{s}\) . For each of these times, relate the direction of the net force to the direction that the airplane is turning, and to whether the airplane is speeding up or slowing down (or neither). (c) What are the magnitude and direction of the net force at \(t=3.00 \mathrm{s} ?\)
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