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A man pulls himself up the \(15^{\circ}\) incline by the method shown. If the combined mass of the man and cart is \(100 \mathrm{kg}\), determine the acceleration of the cart if the man exerts a pull of \(175 \mathrm{N}\) on the rope. Neglect all friction and the mass of the rope, pulleys, and wheels.

Short Answer

Expert verified
The cart's acceleration is approximately 0.338 m/s².

Step by step solution

01

Analyze the Forces

The man pulls the rope with a force of 175 N. This force causes the cart to accelerate up the incline. We consider the gravitational force acting on the mass parallel to the incline, which is the component that opposes the motion. The gravitational force parallel to the incline is \(mg \sin(\theta)\).
02

Calculate Gravitational Force Component

The component of the gravitational force acting parallel to the incline is calculated as \((100 \text{ kg}) \times 9.8 \text{ m/s}^2 \times \sin(15^{\circ})\). Compute this value.
03

Set Up Newton's Second Law

According to Newton's second law, the net force acting on the cart equals the product of its mass and acceleration: \(F_{\text{net}} = ma\). The net force here is the difference between the pull force and the gravitational component: \(175 - mg \sin(15^{\circ}) = ma\).
04

Solve for Acceleration

Rearrange the equation from Step 3 to solve for the acceleration \((a)\). This is done by dividing both sides by the mass (100 kg). Compute the value \(a = \frac{F_{\text{pull}} - mg \sin(15^{\circ})}{m}\).
05

Compute Acceleration

Calculate the final value of acceleration by substituting back the numbers: \(a = \frac{175 \text{ N} - 100 \times 9.8 \times \sin(15^{\circ})}{100}\). Evaluate this expression to find the 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 a cornerstone of classical mechanics. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration. This can be mathematically expressed as \( F = ma \), where \( F \) is the force in newtons (N), \( m \) is the mass in kilograms (kg), and \( a \) is the acceleration in meters per second squared (m/s²).

In practical terms, this means that any change in an object's motion (such as speeding up or slowing down) requires a force to be applied. The greater the mass of an object, the more force is needed to change its motion. In the context of the exercise, the man uses his force to create an acceleration in the cart which includes both his weight and the cart’s weight.

Understanding Newton's Second Law helps us predict how an object will move when subjected to various forces. It's essential in analyzing any moving system, from a simple cart on a track to complex machinery. Whether dealing with inclined planes, rockets, or vehicles, this law lets us calculate the necessary force to achieve desired motion.
Inclined Planes
Inclined planes are flat surfaces tilted at an angle to the horizontal, and they present unique challenges and benefits in force and motion analysis. When dealing with inclined planes, one of the key tasks is to break down forces into components parallel and perpendicular to the incline.

For example, the gravitational force acting on an object on an inclined plane can be split into two components:
  • Parallel to the plane: This component is responsible for pulling the object down the incline and is calculated by \( mg \sin(\theta) \), where \( \theta \) is the angle of the incline, \( m \) is the mass, and \( g \) is the acceleration due to gravity.
  • Perpendicular to the plane: This does not affect the motion along the plane but contributes to the normal force.
In the exercise context, the 15-degree incline affects how the man's pull impacts the cart's motion, modifying how force translates into motion.

Understanding inclined planes allows us to create solutions for moving objects efficiently where elevators or vertical lifts might not be practical. They are used in ramps, slides, and even roads on hills for energy-efficient transport and work.
Force Analysis
Force analysis involves identifying and examining all the forces acting upon an object to understand how they influence its motion. This process often requires breaking down forces into components, especially in systems involving different directions—like on an inclined plane.

In our exercise, the man's pulling force and gravitational forces are the most significant. The steps in the force analysis include:
  • Determine the force applied by the man: Known as the applied force or pull, it's 175 N in this instance.
  • Identify the gravitational component pulling down the slope: Calculated as \( 100 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(15^{\circ}) \). This component opposes the pulling force.
  • Calculate the net force: This is essential for finding acceleration. It's the difference between the applied force and the gravitational pull on the incline.
By solving the net force from these contributors, one can determine the resulting motion of the system using Newton's Second Law.

Force analysis is crucial as it lets us predict and manipulate how objects will react to different force combinations, whether in engineering, design, or basic physics problems.

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Most popular questions from this chapter

The motor unit \(A\) is used to elevate the 300 -kg cylinder at a constant rate of \(2 \mathrm{m} / \mathrm{s}\). If the power meter \(B\) registers an electrical input of \(2.20 \mathrm{kW}\) calculate the combined electrical and mechanical efficiency \(e\) of the system.

The spring of stiffness \(k\) is compressed and suddenly released, sending the particle of mass \(m\) sliding along the track. Determine the minimum spring compression \(\delta\) for which the particle will not lose contact with the loop-the-loop track. The sliding surface is smooth except for the rough portion of length \(s\) equal to \(R,\) where the coefficient of kinetic friction is \(\mu_{k}\).

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Beginning from rest when \(\theta=20^{\circ},\) a 35 -kg child slides with negligible friction down the sliding board which is in the shape of a 2.5 -m circular arc. Determine the tangential acceleration and speed of the child, and the normal force exerted on her \((a)\) when \(\theta=30^{\circ}\) and \((b)\) when \(\theta=90^{\circ}\).

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