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Chapter 4: Problem 78

An object (mass \(3.0 \mathrm{~kg}\) ) slides upward on a vertical wall at constant velocity when a force \(F\) of \(60 \mathrm{~N}\) acts on it at an angle of \(60^{\circ}\) to the horizontal. (a) Draw the freebody diagram of the object. (b) Using Newton's laws find the normal force on the object. (c) Determine the force of kinetic friction on the object.

Short Answer

Expert verified
Normal force: 30 N, Kinetic friction: 22.56 N.

Step by step solution

01

Understand the given information

We need to analyze a scenario where a box of mass 3.0 kg is sliding up a vertical wall with constant velocity. A force \( F = 60 \text{ N} \) is applied at an angle of \( 60^{\circ} \) to the horizontal.
02

Free Body Diagram

In the free body diagram, we identify the forces acting on the object:1. The gravitational force \( \vec{F}_g \) acting downward, \( F_g = mg = 3.0 \times 9.8 = 29.4 \text{ N} \).2. The applied force \( \vec{F} \) with components \( F \cos 60^{\circ} \) horizontally and \( F \sin 60^{\circ} \) vertically.3. The normal force \( \vec{N} \) acting horizontally.4. The kinetic friction force \( \vec{f}_k \) acting downwards along the wall.
03

Resolve the applied force

Resolve the force \( \vec{F} \) into horizontal and vertical components:- Horizontal component: \( F_x = 60 \cos 60^{\circ} = 30 \text{ N} \)- Vertical component: \( F_y = 60 \sin 60^{\circ} = 51.96 \text{ N} \)
04

Apply Newton's Second Law in the vertical direction

For constant velocity, the acceleration is zero, so net force vertically is zero:\( F_y - f_k - F_g = 0 \)Substitute the known values:\( 51.96 - f_k - 29.4 = 0 \)
05

Apply Newton's Second Law in the horizontal direction

In the horizontal direction, since the object does not move into or away from the wall, the net force is zero:\( N - F_x = 0 \)So, \( N = F_x = 30 \text{ N} \)
06

Solve for the force of kinetic friction

Solve the equation from Step 4 for \( f_k \):\( f_k = 51.96 - 29.4 = 22.56 \text{ N} \)
07

Compile Results

Based on the Newton's Second Law calculations:- The normal force on the object is \( 30 \text{ N} \).- The force of kinetic friction is \( 22.56 \text{ N} \).

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Key Concepts

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

Free Body Diagram
When analyzing a physics problem involving multiple forces, like an object moving up a wall, the free body diagram is an essential starting point. It visually represents all the forces acting on the object.
For the given exercise, let's break down the forces:
  • Gravitational Force: This force acts downward due to gravity and is represented as \( \vec{F}_g \). For our 3.0 kg mass, this force is \( F_g = mg = 29.4 \text{ N}\) downward.
  • Applied Force: This is the force causing motion. It's given as \( 60 \text{ N} \) at a \( 60^{\circ} \) angle to the horizontal. It has two components:
    • Horizontal Component: \( F_x = F \cos 60^{\circ} = 30 \text{ N} \)
    • Vertical Component: \( F_y = F \sin 60^{\circ} = 51.96 \text{ N} \)
  • Normal Force: This force is exerted by the wall on the object, acting perpendicular to the wall.
  • Kinetic Friction: Opposes the object's movement up the wall and acts downward along the wall surface.
These components are vectors and should be accurately represented in the diagram to ensure a complete analysis of the forces.
Kinetic Friction
Kinetic friction is the force that opposes the movement of two surfaces sliding past one another. It acts in the direction opposite to the motion.
For our scenario where a box slides upward on a wall:
  • The kinetic friction, \( \vec{f}_k \), is acting downward, opposing the upward motion caused by the applied force.
  • The magnitude of kinetic friction can be determined using Newton's second law. In this exercise, we found it to be \( 22.56 \text{ N} \).
This frictional force plays a crucial role as it affects the required applied force needed to maintain constant velocity. Often confused with static friction, note that kinetic friction applies only when there is motion. Recognizing its impact helps us understand why certain amounts of force are needed to slide objects along surfaces.
Normal Force
The normal force is a key component in understanding how objects interact with surfaces. It acts perpendicular to the surface to counterbalance the object's weight and any other forces pressing against the surface.
In the context of the object sliding up a wall:
  • The normal force, \( \vec{N} \), acts horizontally from the wall back towards the object.
  • Using the horizontal equation of motion \( N - F_x = 0 \), we determined that the normal force is equal to the horizontal component of the applied force, which is \( 30 \text{ N} \).
Understanding the normal force is vital as it influences the amount of frictional force present. It's not just counteracting gravity; it also adjusts based on forces applied, such as the horizontal component of the applied force in this exercise. This makes it a crucial factor in determining the motion and equilibrium of objects on surfaces.

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Most popular questions from this chapter

The weight of a 500 -kg object is 4900 N. (a) When the object is on a moving elevator, its measured weight could be (1) zero, (2) between zero and \(4900 \mathrm{~N}\), (3) more than \(4900 \mathrm{~N},\) (4) all of the preceding. Why? (b) Describe the motion if the object's measured weight is only \(4000 \mathrm{~N}\) in a moving elevator.

A 2.0 -kg object has an acceleration of \(1.5 \mathrm{~m} / \mathrm{s}^{2}\) at \(30^{\circ}\) above the \(-x\) -axis. Write the force vector producing this acceleration in component form.

A horizontal force of \(40 \mathrm{~N}\) acting on a block on a frictionless, level surface produces an acceleration of \(2.5 \mathrm{~m} / \mathrm{s}^{2} .\) A second block, with a mass of \(4.0 \mathrm{~kg}\), is dropped onto the first. What is the magnitude of the acceleration of the combination of blocks if the same force continues to act? (Assume that the second block does not slide on the first block.)

A rope is fixed at both ends on two trees and a bag is hung in the middle of the rope (causing the rope to sag vertically). (a) The tension in the rope depends on (1) only the tree separation, (2) only the sag, (3) both the tree separation and \(\operatorname{sag},(4)\) neither the tree separation nor the sag. \((b)\) If the tree separation is \(10 \mathrm{~m},\) the mass of the bag is \(5.0 \mathrm{~kg},\) and the sag is \(0.20 \mathrm{~m}\), what is the tension in the line?

(a) A horizontal force acts on an object on a frictionless horizontal surface. If the force is halved and the mass of the object is doubled, the acceleration will be (1) four times, (2) two times, (3) one-half, (4) one-fourth as great. (b) If the acceleration of the object is \(1.0 \mathrm{~m} / \mathrm{s}^{2},\) and the force on it is doubled and its mass is halved, what is the new acceleration?
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