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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
1.748 m/s虏

Step by step solution

01

Break Down the Forces

First, break each force into its respective horizontal and vertical components based on the given directions. For the 9.0 N force due east, the entire force acts horizontally, so there are no vertical components. For the 8.0 N force, use trigonometric functions to find the components: horizontal component = \(-8.0 \cos 62^{\circ}\) and vertical component = \(8.0 \sin 62^{\circ}\).
02

Calculate the Net Force

Calculate the net force by adding the horizontal components of the forces and ignoring vertical components, as they do not affect horizontal movement on a frictionless surface. The horizontal component of the 9.0 N force is 9.0 N. The horizontal component of the 8.0 N force, acting westward, is \(-8.0 \cos 62^{\circ}\). Adding these gives: \(F_x = 9.0 - 8.0 \times \cos 62^{\circ}\).
03

Simplify the Force Equation

Simplify by calculating \(\cos 62^{\circ}\) and solving: \(F_x = 9.0 - 8.0 \times 0.4695 \approx 9.0 - 3.756 = 5.244\,\text{N}\). Therefore, the net horizontal force \(F_x\) on the body is approximately 5.244 N acting to the east.
04

Apply Newton's Second Law

Use Newton's second law of motion, \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. Rearrange to find the acceleration: \(a = \frac{F}{m}\). Substitute the values: \(a = \frac{5.244}{3.0}\,\text{m/s}^2\).
05

Calculate the Magnitude of Acceleration

Calculate the acceleration by dividing: \(a \approx \frac{5.244}{3.0} = 1.748\,\text{m/s}^2\). Thus, the magnitude of the body's acceleration is approximately \(1.748\,\text{m/s}^2\).

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Key Concepts

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

Force Components
To understand forces acting on an object, it's useful to break them into their components. When a force acts in a certain direction, it can influence motion in multiple axes.
For example, consider a force acting not purely horizontal or vertical but at an angle. This force can be split into components:
  • Horizontal component: This is how much of the force contributes to movement along the horizontal plane. It's typically calculated using the cosine of the angle.
  • Vertical component: This is how much of the force contributes to movement along the vertical plane. It's calculated using the sine of the angle.
For a force of 8.0 N acting at a 62掳 angle north of west, the horizontal component can be calculated using \\( F_{x} = 8.0 \cos(62^{\circ}) \)\, and the vertical component using \\( F_{y} = 8.0 \sin(62^{\circ}) \).
By determining these components, we gain insights into how the body moves across different directions.
Vector Addition
Once forces are broken into components, the next step is to combine them to find the net effect on the object. This is done using vector addition.
This involves:
  • Adding forces acting in the same direction.
  • Subtracting forces that act in opposite directions.
In the case where two forces act on an object, like 9.0 N due east and an 8.0 N force at an angle, we only consider the horizontal components, because the forces are horizontal, and the surface is frictionless. Forces in the east direction are positive and those in the west are negative.
So, for these forces:
  • The net horizontal force is the sum of the 9.0 N eastward force and the westward component \\(-8.0 \cos(62^{\circ}) \).
This results in a simplified net force, showcasing the overall effect of all horizontal forces.
Acceleration Calculation
After determining the net force, we calculate the acceleration of the object using Newton's second law of motion. This principle states that force is equal to mass times acceleration, or \\( F = ma \).
Rearranging gives us \\( a = \frac{F}{m} \). For the given body of mass 3.0 kg, and knowing the net force to be approximately 5.244 N from our vector addition, the acceleration is calculated by:
  • Substituting these values into the formula: \\[ a = \frac{5.244}{3.0} \approx 1.748 \, \text{m/s}^2 \]
The result is the magnitude of acceleration, indicating how quickly the object moves in response to the applied forces. This method helps in predicting motion accurately and is key in solving physics problems related to dynamics.

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Most popular questions from this chapter

An \(80 \mathrm{~kg}\) man drops to a concrete patio from a window \(0.50 \mathrm{~m}\) above the patio. He neglects to bend his knees on landing, taking \(2.0 \mathrm{~cm}\) to stop. (a) What is his average acceleration from when his feet first touch the patio to when he stops? (b) What is the magnitude of the average stopping force exerted on him by the patio?

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 \(40 \mathrm{~kg}\) girl and an \(8.4 \mathrm{~kg}\) sled are on the frictionless ice of a frozen lake, \(15 \mathrm{~m}\) apart but connected by a rope of negligible mass. The girl exerts a horizontal \(5.2 \mathrm{~N}\) force on the rope. What are the acceleration magnitudes of (a) the sled and (b) the girl? (c) How far from the girl's initial position do they meet?

A \(500 \mathrm{~kg}\) rocket sled can be accelerated at a constant rate from rest to \(1600 \mathrm{~km} / \mathrm{h}\) in \(1.8 \mathrm{~s}\). What is the magnitude of the required net force?

An interstellar ship has a mass of \(1.20 \times 10^{6} \mathrm{~kg}\) and is initially at rest relative to a star system. (a) What constant acceleration is needed to bring the ship up to a speed of \(0.10 c\) (where \(c\) is the speed of light, \(\left.3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)\) relative to the star system in 3.0 days? (b) What is that acceleration in \(g\) units? (c) What force is required for the acceleration? (d) If the engines are shut down when \(0.10 c\) is reached (the speed then remains constant), how long does the ship take (start to finish) to journey 5.0 light-months, the distance that light travels in 5.0 months?
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