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Chapter 10: Problem 16

A \(12.0 \mathrm{~kg}\) box resting on a horizontal, frictionless surface is attached to a \(5.00 \mathrm{~kg}\) weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass \(2.00 \mathrm{~kg}\) and diameter \(0.500 \mathrm{~m} .\) After the system is released, find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force that the axle exerts on the pulley.

Short Answer

Expert verified
The tension in the wire on the side of the 5.00 kg weight is 49.05 N and the side of the 12.0 kg box is 117.72 N. The acceleration of the box is 5.72 m/s^2. The horizontal and vertical components of the force that the axle exerts on the pulley are 83.385 N and 19.62 N respectively.

Step by step solution

01

Determine the tension on both sides of the pulley

Since there's no friction or air resistance and we assume the wire to be light (its mass doesn't figure into our calculations), the weight on one side of the pulley will exert a force equal to \(5.00 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} = 49.05 \mathrm{~N}\). This is the tension force \(T_1\) that the wire exerts on the pulley and on the box. Likewise, on the other side of the pulley, the box will exert a force equal to \(12.0 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} = 117.72 \mathrm{~N}\). This is the tension force \(T_2\) that counteracts \(T_1\).
02

Find the acceleration of the box

The net force on the box is the difference between \(T_1\) and \(T_2\), which is \(117.72 \mathrm{~N} - 49.05 \mathrm{~N} = 68.67 \mathrm{~N}\). However, the net force is also equal to the mass of the box times its acceleration \(a\) (\(F=ma\)). Thus, the acceleration of the box is \(a = \frac{F}{m} = \frac{68.67 \mathrm{~N}}{12.0 \mathrm{~kg}} = 5.72 \mathrm{~m/s^2}\).
03

Calculate the forces on the pulley

The horizontal component of the force that the axle exerts on the pulley is the sum of the tensions \(T_1\) and \(T_2\) divided by 2, which is \(\frac{49.05 \mathrm{~N} + 117.72 \mathrm{~N}}{2} = 83.385 \mathrm{~N}\). The vertical component is simply the weight of the pulley, which is \(2.00 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} = 19.62 \mathrm{~N}\).

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Key Concepts

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

Newton's Laws of Motion
Newton's Laws of Motion form the foundation of classical mechanics. These laws describe how things move and interact. The first law, often called the law of inertia, states that an object at rest stays at rest, and an object in motion stays in motion, unless acted upon by an external force.
The second law defines the relationship between force, mass, and acceleration as \( F = ma \). This means the acceleration of an object is directly proportional to the force acting upon it and inversely proportional to its mass.
The third law is often summarized as "for every action, there is an equal and opposite reaction." This means forces come in pairs. When one body exerts a force on a second body, the second body exerts an equal and opposite force on the first body.
In our problem, these laws apply when calculating the tension and forces acting on the box and pulley system. The second law helps us understand how different forces result in acceleration, while the third law shows the interaction between the box, wire, and pulley.
Tension in Physics
Tension is a force experienced by a rope, string, or wire when it is pulled tight by forces acting from opposite ends. In physics, tension is a pulling force that acts along the length of a wire or rope.
  • When dealing with tension, it is important to remember that it is a vector. This means it has both magnitude and direction.
  • In our system, the tension in the wire connects the box and the weight through the pulley.
  • Tension is crucial for understanding how forces are transmitted between connected objects.
The exercise illustrates tension on both sides of a pulley. We calculated these as \( T_1 = 49.05 \, \text{N} \) and \( T_2 = 117.72 \, \text{N} \). These forces counteract each other, working against gravity in part of the system. Tensions are responsible for conveying forces through the system, affecting the resulting motion of the block and weight.
Rotational Dynamics
Rotational dynamics extends Newton's Laws to rotational motion, covering angular quantities like torque and angular acceleration. For a rotating object, torque replaces force, and moment of inertia replaces mass.
  • Torque, the rotational equivalent of force, is the product of force and the distance from the rotation axis ( \( \tau = r \times F \).
  • The moment of inertia ( \( I \) ) represents an object's resistance to changes in rotational motion.
  • In our pulley system, the pulley rotates due to the unequal tension on either side of its axle.
The net torque on the pulley determines its angular acceleration, solving real-world problems by relating linear and rotational dynamics.
Pulley Systems
Pulley systems make use of wheels and ropes to change the direction and magnitude of forces, often making it easier to lift heavy loads or transfer force. Pulleys are integral for understanding mechanics in physics.
  • A simple pulley consists of a wheel on an axle and helps redirect the force applied.
  • In physics exercises, pulleys are often idealized as frictionless with a light, mass-less rope.
In the context of our problem, the pulley system allows us to connect a box and a weight through a wire, using energy and forces to move the box on a horizontal plane and lift the weight vertically. This demonstrates how pulley systems change the direction of forces to solve mechanical problems efficiently.

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Most popular questions from this chapter

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3 ). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\). If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{\mathrm{k}}\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0\) Rolling without slipping sets in when \(v_{x}=r \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

You are designing a system for moving aluminum cylinders from the ground to a loading dock. You use a sturdy wooden ramp that is \(6.00 \mathrm{~m}\) long and inclined at \(37.0^{\circ}\) above the horizontal. Each cylinder is fitted with a light, frictionless yoke through its center, and a light (but strong) rope is attached to the yoke. Each cylinder is uniform and has mass 460 \(\mathrm{kg}\) and radius \(0.300 \mathrm{~m} .\) The cylinders are pulled up the ramp by applying a constant force \(\overrightarrow{\boldsymbol{F}}\) to the free end of the rope. \(\overrightarrow{\boldsymbol{F}}\) is parallel to the surface of the ramp and exerts no torque on the cylinder. The coefficient of static friction between the ramp surface and the cylinder is \(0.120 .\) (a) What is the largest magnitude \(\overrightarrow{\boldsymbol{F}}\) can have so that the cylinder still rolls without slipping as it moves up the ramp? (b) If the cylinder starts from rest at the bottom of the ramp and rolls without slipping as it moves up the ramp, what is the shortest time it can take the cylinder to reach the top of the ramp?

A thin-walled, hollow spherical shell of mass \(m\) and radius \(r\) starts from rest and rolls without slipping down a track (Fig. \(\mathbf{P 1 0 . 7 2}\) ). Points \(A\) and \(B\) are on a circular part of the track having radius \(R\). The diameter of the shell is very small compared to \(h_{0}\) and \(R,\) and the work done by rolling friction is negligible. (a) What is the minimum height \(h_{0}\) for which this shell will make a complete loop-the-loop on the circular part of the track? (b) How hard does the track push on the shell at point \(B,\) which is at the same level as the center of the circle? (c) Suppose that the track had no friction and the shell was released from the same height \(h_{0}\) you found in part (a). Would it make a complete loop-theloop? How do you know? (d) In part (c), how hard does the track push on the shell at point \(A,\) the top of the circle? How hard did it push on the shell in part (a)?

A metal bar is in the \(x y\) -plane with one end of the bar at the origin. A force \(\overrightarrow{\boldsymbol{F}}=(7.00 \mathrm{~N}) \hat{\imath}+(-3.00 \mathrm{~N}) \hat{\jmath}\) is applied to the bar at the point \(x=3.00 \mathrm{~m}, y=4.00 \mathrm{~m}\). (a) In terms of unit vectors \(\hat{\imath}\) and \(\hat{\jmath},\) what is the position vector \(\vec{r}\) for the point where the force is applied? (b) What are the magnitude and direction of the torque with respect to the origin produced by \(\overrightarrow{\boldsymbol{F}} ?\)

If the body's center of mass were not placed on the rotational axis of the turntable, how would the person's measured moment of inertia compare to the moment of inertia for rotation about the center of mass? (a) The measured moment of inertia would be too large; (b) the measured moment of inertia would be too small; (c) the two moments of inertia would be the same; (d) it depends on where the body's center of mass is placed relative to the center of the turntable.
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