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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 \( 2.93 \text{ m/s}^2 \).

Step by step solution

01

Resolve the Forces into Components

First, resolve each force into its components. The 9.0 N force acting due east has components: \( F_{x_1} = 9.0 \text{ N} \) and \( F_{y_1} = 0 \text{ N} \). The 8.0 N force acting 62掳 north of west has components:\[ F_{x_2} = 8.0 \cos(180掳 - 62掳) \]\[ F_{x_2} = 8.0 \cos(118掳) \approx -3.75 \text{ N} \]\[ F_{y_2} = 8.0 \sin(118掳) \approx 7.06 \text{ N} \].
02

Calculate Net Force Components

Calculate the net force components:For the x-component:\[ F_{x_{ ext{net}}} = F_{x_1} + F_{x_2} = 9.0 + (-3.75) = 5.25 \text{ N} \]For the y-component:\[ F_{y_{ ext{net}}} = F_{y_1} + F_{y_2} = 0 + 7.06 = 7.06 \text{ N} \].
03

Determine the Magnitude of the Net Force

Use the Pythagorean theorem to find the magnitude of the net force vector:\[ F_{ ext{net}} = \sqrt{(F_{x_{ ext{net}}})^2 + (F_{y_{ ext{net}}})^2} \]\[ F_{ ext{net}} = \sqrt{(5.25)^2 + (7.06)^2} \approx \sqrt{27.56 + 49.80} \approx \sqrt{77.36} \approx 8.79 \text{ N} \].
04

Calculate the Acceleration

Use Newton's second law to calculate acceleration:The formula for acceleration is \( a = \frac{F_{ ext{net}}}{m} \), where \( m \) is the mass of the body.\[ a = \frac{8.79 \text{ N}}{3.0 \text{ kg}} \approx 2.93 \text{ m/s}^2 \].

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Key Concepts

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

Force resolution
When forces act at angles, it's essential to resolve each into horizontal and vertical components. This process is known as force resolution. Given a force and its angle relative to a reference direction, such as the east or north, we use trigonometric functions to find components.
  • The cosine function helps find the horizontal component: for a force \( F \) at angle \( \theta \) from the east, the eastward component is \( F \cos \theta \). For north of west, be careful with angles, often involving transformations like \(180^\circ - \theta \).
  • The sine function determines the vertical component: \( F \sin \theta \).
Breaking forces into components allows us to analyze their effects more easily in each direction. This skill is crucial for solving more advanced physics problems, especially when multiple forces are acting.
Net force calculation
After resolving forces into components, the next step is calculating the net force. This process involves finding the sum of forces acting in each direction.
  • Add all horizontal components together to get the net horizontal force.
  • Similarly, sum all vertical components to determine the net vertical force.
The net force in any direction is simply the vector sum of individual forces in that direction. Be mindful of the signs, as they indicate the direction of force: positive in conventional east or north directions and negative otherwise. This method allows us to accurately predict the body's response, as given by Newton's second law, to the total or net force acting on it.
Acceleration calculation
Newton's Second Law of Motion states that a body accelerates in response to a net force acting upon it, following the formula: \( a = \frac{F_{\text{net}}}{m} \). To find acceleration:
  • First, determine the net force on the body using previously calculated net force components.
  • Then, apply the law by dividing this net force by the object's mass.
In our example, with the net force of approximately \(8.79 \text{ N}\) and a mass of \(3.0 \text{ kg}\), the acceleration comes out to about \(2.93 \text{ m/s}^2\). Understanding this process is key, as it links how forces change an object's motion, which is central in dynamics.
Pythagorean theorem in physics
The Pythagorean theorem often appears in physics for calculating resultant magnitudes from component vectors. Given net force components:
  • The theorem states the magnitude of the resultant vector is the square root of the sum of the squares of its components.
For forces, this means: if \( F_{x_{\text{net}}} \) and \( F_{y_{\text{net}}} \) are the net force components, then the net force magnitude \( F_{\text{net}} \) is \( \sqrt{(F_{x_{\text{net}}})^2 + (F_{y_{\text{net}}})^2} \).
In practice, this theorem helps transform complex vector sums into a simple scalar quantity. This is incredibly useful, particularly when dealing with force vectors at angles, turning geometry into straightforward algebra.

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Most popular questions from this chapter

Three forces act on a particle that moves with unchanging velocity \(\vec{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(7 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). Two of the forces are \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((3 \mathrm{~N}) \hat{\mathrm{j}}+(-2 \mathrm{~N}) \hat{\mathrm{k}}\) and \(\vec{F}_{2}=(-5 \mathrm{~N}) \hat{\mathrm{i}}+(8 \mathrm{~N}) \hat{\mathrm{j}}+(-2 \mathrm{~N}) \hat{\mathrm{k}}\). What is the third force?

In 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.

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?

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, \(3.0 \times 10^{8} \mathrm{~m} / \mathrm{s}\) ) 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?

An \(85 \mathrm{~kg}\) man lowers himself to the ground from a height of \(10.0 \mathrm{~m}\) by holding onto a rope that runs over a frictionless pulley to a \(65 \mathrm{~kg}\) sandbag. With what speed does the man hit the ground if he started from rest?
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