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Chapter 5: Problem 1

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?

Short Answer

Expert verified
The magnitude of the body's acceleration is approximately 2.93 m/s虏.

Step by step solution

01

Break Down the Forces

First, we need to break down the forces into their components. The force due east is already aligned with the coordinate axis. Therefore, the force has components:\[F_{1x} = 9.0 \, \text{N}, \quad F_{1y} = 0 \, \text{N}\]The force of 8.0 N acting north of west needs to be broken into its components. Using the angle of \(62^\circ\), \[F_{2x} = -8.0 \, \text{N} \times \cos(62^\circ), \quad F_{2y} = 8.0 \, \text{N} \times \sin(62^\circ)\]
02

Calculate Force Components

Now, calculate the numerical values for each component using trigonometric functions.\[F_{2x} = -8.0 \times \cos(62^\circ) \approx -3.76 \, \text{N}\]\[F_{2y} = 8.0 \times \sin(62^\circ) \approx 7.08 \, \text{N}\]
03

Sum of Forces in Each Direction

Find the net force in each direction by adding the components.For the x-direction:\[F_{x\text{ net}} = F_{1x} + F_{2x} = 9.0 + (-3.76) = 5.24 \, \text{N}\]For the y-direction:\[F_{y\text{ net}} = F_{1y} + F_{2y} = 0 + 7.08 = 7.08 \, \text{N}\]
04

Calculate the Net Force Magnitude

Use the Pythagorean theorem to find the magnitude of the net force:\[F_{\text{net}} = \sqrt{F_{x\text{ net}}^2 + F_{y\text{ net}}^2}\]Substitute the values:\[F_{\text{net}} = \sqrt{(5.24)^2 + (7.08)^2} \approx 8.78 \, \text{N}\]
05

Determine the Acceleration

Use Newton's second law of motion to determine the acceleration:\[F = ma \Rightarrow a = \frac{F}{m}\]Substitute in the known values:\[a = \frac{8.78}{3.0} \approx 2.93 \, \text{m/s}^2\]

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Key Concepts

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

Force Components
When dealing with multiple forces acting on an object, breaking them down into components makes solving the problem more manageable. Each force can be split into parts that act along the x and y directions.
For example, if a force acts due east, like our 9.0 N force, it aligns directly with the x-axis, which means:
  • The x-component is the same as the force itself, here it's 9.0 N.
  • The y-component is zero since the force is not acting in the vertical plane.

On the other hand, for a force acting at an angle, such as the 8.0 N force north of west, you need to use reference angles to determine its x and y components. This makes it essential to understand angles and how to apply them effectively in physics problems.
Net Force Calculation
Calculating the net force involves adding all the force components along each axis. We begin by finding the sum of forces in both the x and y directions, which will tell us how the object behaves under different influences.
In our exercise:
  • The net force in the x-direction is simply the sum of all x-components: 9.0 N and -3.76 N.
  • Adding these results gives us a net x-force of 5.24 N.
  • For the y-direction, you only add the y-components, where in this example it was 0 N and 7.08 N, which equals a net y-force of 7.08 N.

Finding these sums allows us to visualize the total or resultant force on the object, which is crucial for determining movement and acceleration using Newton's second law.
Trigonometric Functions
Trigonometric functions are essential tools for resolving a force acting at an angle into its components. In physics, these functions help us understand relationships in right triangles, especially with forces.
In this problem:
  • We used cosine to find the x-component of the 8.0 N force. Cosine helps us determine how much of a force is directed along the x-axis, calculated as \(-8.0 \cos(62^\circ)\approx -3.76 \, \text{N}\).
  • For the y-component, sine gives us the amount acting vertically, calculated as \(8.0 \sin(62^\circ)\approx 7.08 \, \text{N}\).

Understanding these trigonometric functions allows us to efficiently resolve forces and analyze their effects on objects.
Pythagorean Theorem
The Pythagorean theorem provides a way to calculate the resultant or net force from its components. It links the components in a right triangle to the magnitude of the hypotenuse, or in our context, the net force.
Given the formula, \(F_{\text{net}} = \sqrt{F_{x\text{ net}}^2 + F_{y\text{ net}}^2}\), we can find how strong the overall force is.
  • In our example, substituting gives: \(F_{\text{net}} = \sqrt{(5.24)^2 + (7.08)^2} \approx 8.78 \, \text{N}.\)

Using the Pythagorean theorem is an essential skill to synthesize the information from various force components into a clear understanding of the object's real-world interaction.

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Most popular questions from this chapter

A block with a weight of \(3.0 \mathrm{~N}\) is at rest on a horizontal surface. A \(1.0 \mathrm{~N}\) upward force is applied to the block by means of an attached vertical string. What are the (a) magnitude and (b) direction of the force of the block on the horizontal surface?

Compute the weight of a \(75 \mathrm{~kg}\) space ranger (a) on Earth, (b) on Mars, where \(g=3.7 \mathrm{~m} / \mathrm{s}^{2}\), and (c) in interplanetary space, where \(g=0 .(\mathrm{d})\) What is the ranger's mass at each location?

A motorcycle and \(60.0 \mathrm{~kg}\) rider accelerate at \(3.0 \mathrm{~m} / \mathrm{s}^{2}\) up a ramp inclined \(10^{\circ}\) above the horizontal. What are the magnitudes of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

A car that weighs \(1.30 \times 10^{4} \mathrm{~N}\) is initially moving at \(40 \mathrm{~km} / \mathrm{h}\) when the brakes are applied and the car is brought to a stop in \(15 \mathrm{~m}\). Assuming the force that stops the car is constant, find (a) the magnitude of that force and (b) the time required for the change in speed. If the initial speed is doubled, and the car experiences the same force during the braking, by what factors are (c) the stopping distance and (d) the stopping time multiplied? (There could be a lesson here about the danger of driving at high speeds.)

In April 1974 , John Massis of Belgium managed to move two passenger railroad cars. He did so by clamping his teeth down on a bit that was attached to the cars with a rope and then leaning backward while pressing his feet against the railway ties. The cars together weighed \(700 \mathrm{kN}\) (about 80 tons). Assume that he pulled with a constant force that was \(2.5\) times his body weight, at an upward angle \(\theta\) of \(30^{\circ}\) from the horizontal. His mass was \(80 \mathrm{~kg}\), and he moved the cars by \(1.0 \mathrm{~m}\). Neglecting any retarding force from the wheel rotation, find the speed of the cars at the end of the pull.
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