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Chapter 4: Problem 53

ssm Three forces act on a moving object. One force has a magnitude of 80.0 N and is directed due north. Another has a magnitude of 60.0 N and is directed due west. What must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?

Short Answer

Expert verified
The third force has a magnitude of 100 N and is directed 53.1掳 south of east.

Step by step solution

01

Understanding Constant Velocity

When an object moves with a constant velocity, it is not accelerating. According to Newton's first law, this means that the net force acting on the object is zero.
02

Identify the Forces as Vectors

We have two forces given: 80.0 N directed due north, represented as a vector (0, 80) N, and 60.0 N directed due west, represented as (-60, 0) N. These vectors will guide us in determining the third force.
03

Calculate the Resultant of Known Forces

The resultant of the two given forces is found by vector addition: \[ \begin{align*} F_{x} &= -60 + 0 = -60 \text{ N} \ F_{y} &= 0 + 80 = 80 \text{ N} \ \end{align*} \]The resultant vector is thus (-60, 80) N.
04

Determine the Third Force

Since the object is moving with constant velocity, the third force must be equal and opposite to the resultant of the first two forces so that they cancel each other: \[ F_{3} = (60, -80) \text{ N} \]
05

Calculate the Magnitude of the Third Force

The magnitude of this third force can be calculated using the Pythagorean theorem: \[ |F_{3}| = \sqrt{(60)^2 + (-80)^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100 \text{ N} \]
06

Determine the Direction of the Third Force

The direction of the force can be found using the arctangent function: \[ \theta = \tan^{-1}\left(\frac{-80}{60}\right) = \tan^{-1}\left(-\frac{4}{3}\right) \]This angle is measured from the positive x-axis (east direction), leading to approximately 53.1 degrees south of east.

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Key Concepts

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

Understanding Constant Velocity
When we discuss constant velocity, we're talking about an object that maintains its speed and direction over time. In simple terms, the object does not speed up, slow down, or change direction. According to Newton's First Law, this happens when the net force acting on the object is zero. That means all the forces acting upon it are perfectly balanced, canceling each other out. For example, if a book is sliding along a surface at a constant velocity with no exerted forces other than gravity and friction, these forces are balanced. No one force overpowers the other, allowing the object to continue its uniform motion.
Breaking Down Vector Addition
Vector addition is the process of combining two or more forces, which are represented as vectors. Each force has both a magnitude (size) and direction. In our exercise, we deal with forces that act in perpendicular directions: one north and another west. When adding these force vectors, we treat them like coordinates on a graph:
  • The northward force becomes a vertical vector on the graph.
  • The westward force is represented as a horizontal vector.
To find the resultant force, imagine these vectors as arrows. We "add" them by stacking them end to end - like laying arrows tip-to-tail. The resultant vector (or net force) emerges from the starting point of the first vector to the end of the last vector. This vector shows us the direction and magnitude of the combined forces.
The Pythagorean Theorem in Force Calculations
The Pythagorean theorem helps us find the magnitude of the resultant force from the vector addition. According to this theorem, in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.
  • Here, the resultant force vector is the hypotenuse.
  • The other sides are the two orthogonal vectors: north and west forces.
  • The formula is: \[ |F_{3}| = \sqrt{(-60)^2 + 80^2} = \sqrt{10000} = 100 \text{ N} \]
This calculation finds the magnitude or size of the new vector created by our two stacked vectors. It tells us the strength of the third force needed to maintain constant velocity.
Using the Arctangent Function to Find Direction
To find the direction of a force vector, we use the arctangent function, often represented as \(\tan^{-1}\) or "inv tan," which calculates angles. It's handy in physics for determining angles in vector problems when we know the opposite and adjacent sides of a right triangle.
  • For our vector, we know the opposite side length as 80 N (north force) and adjacent side length as 60 N (west force).
  • We calculate the angle \[\theta = \tan^{-1}\left(\frac{-80}{60}\right)\]This angle tells us how far we need to rotate from the east direction (positive x-axis) to align with our third force vector.
  • The result, approximately 53.1 degrees, places the force south of east, completing the balance to maintain constant velocity.
Understanding these vector directions is crucial for determining how forces act on moving objects.

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Most popular questions from this chapter

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is 109 kg, and the coefficient of kinetic friction between the road and him is 0.870. Find the tension in the cable.

A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earth and the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m},\) and the mass of the earth is 81.4 times as great as that of the moon.

mmh The drawing shows a large cube (mass \(=25 \mathrm{kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\overrightarrow{\mathbf{P}}\) . A small cube (mass \(=4.0 \mathrm{kg} )\) is in contact with the front surface of the large cube and will slide downward unless \(\overrightarrow{\mathbf{P}}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71 .\) What is the smallest magnitude that \(\overrightarrow{\mathbf{P}}\) can have in order to keep the small cube from sliding downward?

At an instant when a soccer ball is in contact with the foot of a player kicking it, the horizontal or \(x\) component of the ball's acceleration is 810 \(\mathrm{m} / \mathrm{s}^{2}\) and the vertical or \(y\) component of its acceleration is 1100 \(\mathrm{m} / \mathrm{s}^{2}\) . The ball's mass is 0.43 \(\mathrm{kg}\) . What is the magnitude of the net force acting on the soccer ball at this instant?

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