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A man pushes on a piano with mass 180 kg; it slides at constant velocity down a ramp that is inclined at 19.0\(^\circ\) above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline and (b) parallel to the floor.

Short Answer

Expert verified
(a) 574.5 N, (b) 608.0 N

Step by step solution

01

Analyze the forces involved

First, identify the forces acting on the piano. The forces are: the gravitational force (weight) acting downward, the normal force exerted by the ramp perpendicular to its surface, and the force exerted by the man. Since the piano slides at constant velocity, the net force along the incline is zero.
02

Calculate gravitational component parallel to incline

The component of the gravitational force acting parallel to the incline is what the man must counteract to maintain a constant velocity. Calculate this force using:\[ F_{\text{gravity, parallel}} = mg \sin \theta \]where \( m = 180 \text{ kg} \) is the mass of the piano, \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( \theta = 19.0^\circ \) is the incline angle. Substituting the values gives:\[ F_{\text{gravity, parallel}} = 180 \times 9.8 \times \sin(19.0^\circ) \approx 574.5 \, \text{N} \]
03

Calculate force when pushing parallel to the incline

When the man pushes parallel to the incline, the force he exerts must be equal to the component of the gravitational force parallel to the incline, hence:\[ F_{\text{man, parallel}} = 574.5 \, \text{N} \]
04

Calculate normal force

Calculate the component of the gravitational force perpendicular to the incline (the normal force). This is given by:\[ F_{\text{gravity, perpendicular}} = mg \cos \theta \]Substitute the known values:\[ F_{\text{gravity, perpendicular}} = 180 \times 9.8 \times \cos(19.0^\circ) \approx 1671.1 \, \text{N} \]
05

Calculate force when pushing parallel to the floor

When the man pushes parallel to the floor, his force has a horizontal component \( F_{\text{man, horizontal}} \) and a component perpendicular to the incline which modifies the normal force. Using geometrical decomposition:\[ F_{\text{man}} \cos(19.0^\circ) = 574.5 \, \text{N} \]Solve for \( F_{\text{man}} \):\[ F_{\text{man}} = \frac{574.5}{\cos(19.0^\circ)} \approx 608.0 \, \text{N} \]

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Key Concepts

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

Inclined Planes
An inclined plane is a flat surface tilted at an angle, other than a right angle, relative to a horizontal surface. It enables us to study the effects of forces in mechanics, particularly how gravity influences objects on a slope. Imagine a ramp or a hill, where the steepness is gauged by the angle with respect to the horizontal. When dealing with inclined planes in physics, we often decompose forces into components parallel and perpendicular to the plane.
  • Parallel component: This is the part of the force that acts directly along the plane’s surface, causing the object to slide down or move up.
  • Perpendicular component: This force acts at a right angle to the surface of the incline, helping us understand the normal force, which keeps the object pressed against the plane.
The inclined plane showcases basic principles of mechanics, such as how forces interact and change based on orientation and movement environments.
Gravitational Force
Gravitational force is the force of attraction between two masses. On Earth, this force pulls objects toward the ground. For an object like a piano on an inclined plane, gravity doesn't pull it just directly downward, but instead, can be split into two useful components when analyzing motion along the incline.
  • Perpendicular to the incline: This component determines the normal force, keeping the object in contact with the surface.
  • Parallel to the incline: This component causes the object to slide down the ramp, inducing motion.
The gravitational force can be calculated using the formula: \( F = mg \), where \( m \) is mass and \( g \) is the acceleration due to gravity, typically \( 9.8 \, \text{m/s}^2 \) on Earth. Understanding gravitational force is crucial to predicting how objects will behave when placed on inclined surfaces.
Constant Velocity
Constant velocity occurs when an object's speed and direction remain consistent over time. In the context of physics and mechanics, this means the forces acting on the object are balanced, resulting in no net force.
If the piano slides down the inclined plane at a constant velocity, the forces parallel to the incline must cancel out, keeping the speed stable. There's no acceleration, so the gravitational force component down the incline is exactly countered by the force exerted by the man pushing the piano.
This balancing act is what lets us calculate how the pushing force must be equal to the parallel gravitational component in order to maintain constant velocity. Thus, constant velocity provides insights into equilibrium on an incline and helps determine required forces to achieve this state.
Force Components
Force components arise from the decomposition of forces into perpendicular axes, usually aligned with the system being analyzed, like an inclined plane. Decomposing forces simplifies analyzing the situation, allowing us to focus on particular directions.
  • Parallel to Incline: This component must be countered by an applied force to hold constant velocity or induce motion. In this example, it's the man's force against gravity that ensures constant speed downhill.
  • Perpendicular (Normal force): This component keeps the object pressed against the incline, counteracting the perpendicular gravitational component.
Understanding force components equips us with the ability to solve complex dynamics problems by splitting forces into manageable vectors. This foundational concept is vital for tackling any mechanics problem involving inclines or other angled surfaces.

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Most popular questions from this chapter

Two crates connected by a rope lie on a horizontal surface (\(\textbf{Fig. E5.37}\)). Crate \(A\) has mass \(m_A\), and crate \(B\) has mass \(m_B\). The coefficient of kinetic friction between each crate and the surface is \(\mu_k\). The crates are pulled to the right at constant velocity by a horizontal force \(F\). Draw one or more free-body diagrams to calculate the following in terms of \(m_A\), \(m_B\), and \(\mu_k\): (a) the magnitude of \(F\) and (b) the tension in the rope connecting the blocks.

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