Problem 15 - College Physics

Question

A man pushes on a piano of mass \(180 \mathrm{~kg}\) so that it slides at a constant velocity of \(12.0 \mathrm{~cm} / \mathrm{s}\) down a ramp that is inclined at \(11.0^{\circ}\) above the horizontal. No appreciable friction is acting on the piano. Calculate the magnitude and direction of this push (a) if the man pushes parallel to the incline, (b) if the man pushes the piano up the plane instead, also at \(12.0 \mathrm{~cm} / \mathrm{s}\) parallel to the incline, and (c) if the man pushes horizontally but still with a speed of \(12.0 \mathrm{~cm} / \mathrm{s}\)

Step-by-Step Solution

Answer

The push is equal to the gravitational component: \(F = mg \sin(11^\circ)\). It is the same for both cases (up or down) since it's constant velocity. Horizontal push is \(F = \frac{mg \sin(11^\circ)}{\cos(11^\circ)}\).

Step 1: Identify Forces

The forces acting on the piano are the gravitational force, the normal force, and the force applied by the man.

Step 2: Calculate Gravitational Force Component Parallel to Incline

The gravitational force pulling the piano down the incline is given by the component of gravity parallel to the incline: \[ F_{g, \, parallel} = mg \sin(\theta) \]where \( m = 180 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), and \( \theta = 11^{\circ} \). Substitute values to find this component.

Step 3: Determine Force Required for Constant Velocity Downhill

Since the piano is sliding at a constant velocity downhill, the force applied by the man must exactly balance the gravitational force component down the incline: \( F_{\text{man}} = F_{g, \, parallel} \)Substitute the calculated value of \( F_{g, \, parallel} \) to find the required force.

Step 4: Calculate Force for Pushing Uphill

When the man pushes the piano uphill, at constant velocity, he must overcome both the gravitational force component and maintain constant velocity: \[ F_{\text{man, uphill}} = mg \sin(\theta) \]Here, \( F_{\text{man, uphill}} = F_{g, \, parallel} \), hence, it is the same as the downhill force component already calculated.

Step 5: Calculate Horizontal Force Component

If the man pushes horizontally, the horizontal component of the force must balance the gravitational component down the incline. The horizontal force required: \[ F_{\text{horizontal}} = \frac{mg \sin(\theta)}{\cos(\theta)} \]Substitute the known values to find this force.

Key Concepts

Inclined Plane Problems

Inclined plane problems are common in physics and help us understand how objects move on a slope. These problems typically involve calculating the forces that act on an object as it moves along a tilted surface or plane. In our exercise, we have a piano sliding down an inclined ramp with a constant velocity. This means the forces acting on the piano are balanced.

To solve problems involving an inclined plane, it is essential to resolve the forces into components parallel and perpendicular to the incline. This is because gravity is the primary force acting on the object, but it affects the object differently when on an incline than on a flat surface.

Understanding inclined plane problems also involves knowing about frictionless surfaces versus surfaces with friction. In this scenario, the ramp is frictionless, simplifying our calculations, as we don't need to account for any resistive forces due to friction.

Forces on an Incline

Forces acting on an object on an inclined plane include gravitational force, normal force, and any applied force. In the case of our piano, there's no friction, which simplifies the situation considerably.

The gravitational force acts straight down towards the Earth's center. However, when analyzing motion on a slope, it's more useful to break this force into components: one parallel to the incline, which pulls the piano down, and one perpendicular to the incline, which is balanced by the normal force.

The applied force by the man is crucial here. Depending on its direction and magnitude, it can either help or counteract the gravitational force component parallel to the incline. Pushing parallel to or up the incline impacts how much force is needed to maintain a constant velocity.

Constant Velocity

Constant velocity means the speed and direction of the moving object don't change over time. For the piano sliding down the ramp, this constant velocity implies that the net force acting on the piano is zero.

This equilibrium is achieved when the man's push precisely counters the gravitational component pulling the piano down the incline. If the man manages to balance these forces perfectly, the piano's speed remains steady, moving without acceleration.

The significance of constant velocity in physics exercises is that it provides a clear criterion for balancing forces. In our problem, since the piano moves neither faster nor slower, the complications of acceleration or deceleration can be ignored when calculating forces.

Gravitational Force Component

On an inclined plane, the gravitational force acting on an object has two components: one parallel to the slope and one perpendicular

The parallel component is crucial as it dictates the motion along the incline. It can be calculated using the sine function: \[ F_{g, \, parallel} = mg \sin(\theta) \]where \( m \) is the mass, \( g \) is the gravitational acceleration (\(9.8 \, \text{m/s}^2\)), and \( \theta \) is the incline angle.

The perpendicular component, which involves the cosine function, \( F_{g, \, perpendicular} = mg \cos(\theta) \), helps determine the normal force exerted by the surface on the object. In our problem, however, the focus is primarily on the parallel component as it directly impacts the motion of the piano down the incline.