Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 \( F_{\text{man}} \):\[ F_{\text{man}} = \frac{574.5}{\cos(19.0^\circ)} \approx 608.0 \, \text{N} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 2540-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by \(v(t) = At + Bt^2\), where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50 m/s\(^2\) at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m>s. (a) Determine \(A\) and \(B\), including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

You are riding in a school bus. As the bus rounds a flat curve at constant speed, a lunch box with mass 0.500 kg, suspended from the ceiling of the bus by a string 1.80 m long, is found to hang at rest relative to the bus when the string makes an angle of 30.0\(^{\circ}\) with the vertical. In this position the lunch box is 50.0 m from the curve's center of curvature. What is the speed \(\upsilon \) of the bus?

A bowling ball weighing 71.2 N 116.0 lb2 is attached to the ceiling by a 3.80-m rope. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is 4.20 m/s. At this instant, what are (a) the acceleration of the bowling ball, in magnitude and direction, and (b) the tension in the rope?

On the ride "Spindletop" at the amusement park Six Flags Over Texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 m. The cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor dropped about 0.5 m. The people remained pinned against the wall without touching the floor. (a) Draw a force diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction was required for the person not to slide downward to the new position of the floor? (c) Does your answer in part (b) depend on the person's mass? (\(Note\): When such a ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 N at the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? (d) What then would be the passenger's apparent weight at the lowest point?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free