91Ó°ÊÓ

The mechanism shown in Fig. \(\mathbf{P} \mathbf{1 0 . 6 4}\) is used to raise a crate of supplies from a ship's hold. The crate has total mass \(50 \mathrm{~kg} .\) A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius \(0.25 \mathrm{~m}\) and moment of inertia \(I=2.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius \(0.12 \mathrm{~m},\) the cylinder turns, and the crate is raised. What magnitude of the force \(\vec{F}\) applied tangentially to the rotating crank is required to raise the crate with an acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2} ?\) (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

Step by step solution

01

Define the Variables

Let's define the variables. Here, \(m\) is the mass of the crate (50 kg), \(r\) is the radius of the cylinder (0.25 m), \(R\) is the radius of the circle the crank handle rotates around (0.12 m), and \(I\) is the moment of inertia of the cylinder (2.9 kg.m²). The crate's acceleration \(a\) is given as 1.40 m/s², and the tangential force applied to the end of the handle, \(F\), is what we need to find.

02

Apply Newton’s second law

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object. In the context of our problem, the net force acting on the system is equal to the weight of the crate (which is pulling the crate down) minus the force that raises the crate (\(F_{up}\)). We can express \(F_{up}\) as \(F_{up} = m \cdot a + m \cdot g\), where \(g\) is gravitational acceleration. The force \(F_{up}\) is transferred by the cylinder, and we can express this force as the torque on the cylinder, which is equal to the cylinder's moment of inertia multiplied by the angular acceleration, divided by the cylinder's radius.

03

Calculate the Torque

The torque \(\tau\) is the rotational equivalent of force, and it equals the torque produced by the force \(F\) at the crank. It could be given as the product of the force \(F\), the radius of the crank, and the angular acceleration, \(\tau = F \cdot R = \frac {I \cdot a} {r}\). Using this relation, we can find the magnitude of force \(F = \frac {I \cdot a} {R \cdot r}\).

04

Calculate the Force

Substituting the given values into the equation for force \(F\), \(F = \frac {2.9 \cdot 1.40} {0.12 \cdot 0.25}\), we calculate the force required to lift the crate with the given acceleration.

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Key Concepts

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

Newton's second law

Newton's second law is fundamental to understanding motion. It defines how the motion of an object changes when forces are applied.
Simply put, the second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. The formula is \( F = m \cdot a \).
For example, if you push a cart, it accelerates based on its mass and the force you apply. In our exercise, it's important for determining the net force needed to lift the crate, considering gravitational forces as well.

• Net force: Difference between upward force (raising the crate) and gravitational pull (weight).
• Formula used here is \( F_{up} = m \cdot a + m \cdot g \), where \( g \) is the acceleration due to gravity (approx. 9.8 m/s²).

This sets the groundwork for calculating the other forces at play around the rotating cylinder.

moment of inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion.
Think of it as the rotational counterpart to mass in linear dynamics. It's crucial when analyzing systems in rotational motion.
For a cylinder like in the exercise, the moment of inertia \( I \) represents how hard it is to change its state of rotation. It depends on the shape of the object and the mass distribution relative to the axis of rotation.

• The formula for the moment of inertia of a solid cylinder is \( I = \frac{1}{2} m r^2 \), but already given here as \( 2.9 \mathrm{~kg} \cdot \mathrm{m}^2 \).
• This value is used to compute the torque required to raise the crate, showing how moment of inertia factors into force calculations.

Understanding the moment of inertia helps in figuring out the rotational dynamics involved in lifting the crate.

torque

Torque is similar to force but in rotational systems. It measures how much a force causes an object to rotate.
In practical terms, when you turn the crank, you're applying torque that causes the cylinder to rotate.
The applied force at a distance (radius) from the pivot point creates rotational motion.

• In our example, torque \( \tau \) can be calculated using \( \tau = F \cdot R \), where \( R \) is the radius of the crank handle.
• The required torque is also linked to the moment of inertia and angular acceleration, using \( \tau = \frac{I \cdot a}{r} \).

This relation helps determine the needed force \( F \) at the crank to achieve the desired acceleration of the crate.
By understanding torque, you can see how linear forces are converted into rotational actions needed to raise objects like the crate in this exercise.