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Chapter 5: Problem 40

A dated box of dates, of mass \(5.00 \mathrm{~kg}\), is sent sliding up a frictionless ramp at an angle of \(\theta\) to the horizontal. Figure \(5-41\) gives, as a function of time \(t,\) the component \(v_{x}\) of the box's velocity along an \(x\) axis that extends directly up the ramp. What is the magnitude of the normal force on the box from the ramp?

Short Answer

Expert verified
The normal force is \( 49.05 \, \text{N} \cdot \cos(\theta) \).

Step by step solution

01

Understand the Normal Force

The problem asks for the normal force on a box sliding up a frictionless ramp. The normal force acts perpendicular to the surface of the ramp and can be found by considering the forces acting on the box due to gravity.
02

Identify the Forces

The box is subjected to gravitational force, which can be divided into two components: one parallel to the ramp (causing it to slide) and one perpendicular to the ramp (which is balanced by the normal force).
03

Calculate the Gravitational Force

The gravitational force acting on the box is given by the equation \[ F_{gravity} = m imes g \]where the mass \( m = 5.00 \, \text{kg} \) and the acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \). Thus,\[ F_{gravity} = 5.00 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 49.05 \, \text{N} \].
04

Decompose Gravitational Force

Next, decompose the gravitational force into components. The component perpendicular to the ramp is \[ F_{gravity, \perp} = F_{gravity} \cdot \cos(\theta) \],because the angle between the gravitational force and its perpendicular component is \(\theta\).
05

Calculate the Normal Force

Since the ramp is frictionless and only the gravitational component perpendicular to the ramp interacts with the normal force, the magnitude of the normal force \( F_N \) is equal to \( F_{gravity, \perp} \).Thus, \[ F_N = 49.05 \, \text{N} \cdot \cos(\theta) \].

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental concept in physics that keeps objects grounded. It acts on any object with mass and pulls it towards the center of the Earth.
This force is defined by the formula:
  • \( F_{gravity} = m \times g \)
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, typically \( 9.81 \, \text{m/s}^2 \).

In the context of our dated box, its mass is \( 5.00 \, \text{kg} \), hence the gravitational force is \( 49.05 \, \text{N} \).
This force can be broken down into components when the box is on a ramp, which leads us to explore how the gravitational force behaves in such scenarios.
Frictionless Ramp
A frictionless ramp is a theoretical concept where surfaces have no resistance to movement. This means that an object can slide along the ramp without any frictional forces slowing it down.
On a frictionless ramp, the only forces acting on the box are gravitational. The absence of friction simplifies calculations and focuses on analyzing the normal force and gravitational components.
  • This environment allows us to neglect any sliding resistance, making the normal and gravitational force analysis cleaner.
In this scenario, the ramp is inclined at an angle \( \theta \) to the horizontal, leading to a decomposition of gravitational force that directs part of it parallel to the ramp's surface.
Components of Force
For a box on an inclined plane, the gravitational force splits into two components:
  • Parallel Component: This drives the object up or down the ramp. It is calculated as \( F_{gravity, \parallel} = F_{gravity} \sin(\theta) \).
  • Perpendicular Component: This acts perpendicular to the surface and is responsible for defining the normal force. It is calculated as \( F_{gravity, \perp} = F_{gravity} \cos(\theta) \).
Understanding these components is crucial. The normal force, which acts to balance the perpendicular component of the gravitational force, is why the box does not sink into the ramp.

In our exercise, the normal force \( F_N \) is equal to the perpendicular gravitational component \( F_{gravity, \perp} \), given by \( 49.05 \, \text{N} \cos(\theta) \).
This simplification is possible due to the frictionless nature of the ramp, which means no other forces are contributing perpendicularly.

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Most popular questions from this chapter

In shot putting, many athletes elect to launch the shot at an angle that is smaller than the theoretical one (about \(42^{\circ}\) ) at which the distance of a projected ball at the same speed and height is greatest. One reason has to do with the speed the athlete can give the shot during the acceleration phase of the throw. Assume that a \(7.260 \mathrm{~kg}\) shot is accelerated along a straight path of length \(1.650 \mathrm{~m}\) by a constant applied force of magnitude \(380.0 \mathrm{~N}\), starting with an initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (due to the athlete's preliminary motion). What is the shot's speed at the end of the acceleration phase if the angle between the path and the horizontal is (a) \(30.00^{\circ}\) and (b) \(42.00^{\circ}\) ? (Hint: Treat the motion as though it were along a ramp at the given angle. \()\) (c) By what percent is the launch speed decreased if the athlete increases the angle from \(30.00^{\circ}\) to \(42.00^{\circ} ?\)

A shot putter launches a \(7.260 \mathrm{~kg}\) shot by pushing it along a straight line of length \(1.650 \mathrm{~m}\) and at an angle of \(34.10^{\circ}\) from the horizontal, accelerating the shot to the launch speed from its initial speed of \(2.500 \mathrm{~m} / \mathrm{s}\) (which is due to the athlete's preliminary motion). The shot leaves the hand at a height of \(2.110 \mathrm{~m}\) and at an angle of \(34.10^{\circ},\) and it lands at a horizontal distance of \(15.90 \mathrm{~m} .\) What is the magnitude of the athlete's average force on the shot during the acceleration phase? (Hint: Treat the motion during the acceleration phase as though it were along a ramp at the given angle.

Only two horizontal forces act on a \(3.0 \mathrm{~kg}\) body that can move over a frictionless floor. One force is \(9.0 \mathrm{~N}\), acting due east, and the other is \(8.0 \mathrm{~N}\), acting \(62^{\circ}\) north of west. What is the magnitude of the body's acceleration?

If the \(1 \mathrm{~kg}\) standard body has an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(20.0^{\circ}\) to the positive direction of an \(x\) axis, what are (a) the \(x\) component and (b) the \(y\) component of the net force acting on the body, and (c) what is the net force in unit-vector notation?

An electron with a speed of \(1.2 \times 10^{7} \mathrm{~m} / \mathrm{s}\) moves horizontally into a region where a constant vertical force of \(4.5 \times 10^{-16} \mathrm{~N}\) acts on it. The mass of the electron is \(9.11 \times 10^{-31} \mathrm{~kg} .\) Determine the vertical distance the electron is deflected during the time it has moved \(30 \mathrm{~mm}\) horizontally.
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