/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 Two ropes are attached to a heav... [FREE SOLUTION] | 91Ó°ÊÓ

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Two ropes are attached to a heavy box to pull it along the floor. One rope applies a force of 475 newtons in a direction due west; the other applies a force of 315 newtons in a direction due south. As we will see later in the text, force is a vector quantity. (a) How much force should be applied by a single rope, and (b) in what direction (relative to due west), if it is to accomplish the same effect as the two forces added together?

Short Answer

Expert verified
Resultant force: 570 N, direction: 33.4° south of west.

Step by step solution

01

Identify Given Values

We have two forces to consider: 475 newtons due west and 315 newtons due south.
02

Utilize the Pythagorean Theorem

The forces are perpendicular to each other, so their combined magnitude can be found using the Pythagorean theorem. Calculate the magnitude of the resultant force vector using: \[ F = \sqrt{F_{west}^2 + F_{south}^2} = \sqrt{475^2 + 315^2} \]
03

Calculate the Magnitude of the Resultant Force

Calculate the magnitude: \[ F = \sqrt{475^2 + 315^2} = \sqrt{225625 + 99225} = \sqrt{324850} \approx 570 \text{ newtons} \]
04

Calculate the Direction of the Resultant Force

The direction relative to due west is found using trigonometry. We use the tangent ratio: \[ \tan(\theta) = \frac{F_{south}}{F_{west}} = \frac{315}{475} \]
05

Calculate the Angle Relative to Due West

Calculate \( \theta \): \[ \theta = \arctan\left(\frac{315}{475}\right) \approx 33.4^\circ \]
06

Express the Direction of the Single Force

Thus, the direction of the single resultant force is 33.4 degrees south of west.

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Key Concepts

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

Resultant Force
When dealing with multiple forces, like in the case of pulling a heavy box along the floor with ropes, you need to determine the single force that can replace the multiple forces. This equivalent single force is known as the **resultant force**. Understanding the resultant force is crucial in physics as multiple forces often act on an object simultaneously, and the resultant force helps in understanding the net effect of these forces on the object's motion.

To find the resultant force when two forces are acting at a 90-degree angle to each other, you combine them using vector addition. The magnitude and direction of this resultant force are the keys to equilibrium. The magnitude tells you the strength of the force, and the direction tells you where that force is aimed.

This concept is essential not just in physics classes, but also in real-life applications like engineering, where structures often have multiple forces acting on them.
Pythagorean Theorem
The **Pythagorean theorem** is a fundamental concept used to solve problems involving right triangles. It states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

Imagine our forces acting at right angles to each other—due west and due south. We can treat these force vectors as the legs of a right triangle, with the resultant force being the hypotenuse. By applying the Pythagorean theorem, it becomes:
  • Force west: 475 newtons
  • Force south: 315 newtons
  • Resultant force: \[ F = \sqrt{475^2 + 315^2} \approx 570 \text{ newtons} \]
This calculation shows how the theorem is applied to find the magnitude of the resultant force. Understanding the Pythagorean theorem is critical when working with perpendicular force vectors.
Force Vectors
In physics, a **force vector** is a quantity that has both magnitude and direction. Unlike scalar quantities, which only have magnitude (like mass or speed), vectors need direction to fully describe them.

Force vectors are crucial because they allow us to understand how multiple forces combine to move an object. In our example, you have one force vector heading due west and another going due south. To find the resultant force, the vectors need to be added together vectorially, considering both magnitude and direction.

The addition of force vectors is done geometrically or algebraically, as shown with the Pythagorean theorem, helping determine the resultant vector that represents the combined effect of the original forces. This is a fundamental skill in physics, as it lays the groundwork for analyzing many real-world problems.
Trigonometry in Physics
**Trigonometry** is a branch of mathematics that deals with the relationships between the sides and angles of triangles. In physics, it is used to resolve forces where we need both their magnitude and direction.

In our case, once the magnitude of the resultant force is determined, trigonometry helps us find the direction. We use the tangent function, which relates the angle to the ratio of the opposite side over the adjacent side in the right triangle formed by the force vectors. Specifically, we calculate the angle as follows:
  • Using tangent: \[ \tan(\theta) = \frac{F_{south}}{F_{west}} \]
  • To find the angle: \[ \theta = \arctan\left(\frac{315}{475}\right) \approx 33.4^\circ \]
This angle gives the direction of the resultant force, which is 33.4 degrees south of west. Mastery of trigonometry is vital in physics for analyzing forces that act at various angles.

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Most popular questions from this chapter

The \(x\) vector component of a displacement vector \(\overrightarrow{\mathbf{r}}\) has a magnitude of \(125 \mathrm{~m}\) and points along the negative \(x\) axis. The \(y\) vector component has a magnitude of \(184 \mathrm{~m}\) and points along the negative \(y\) axis. Find the magnitude and direction of \(\overrightarrow{\mathbf{r}}\). Specify the direction with respect to the negative \(x\) axis.

Before starting this problem, review Interactive Solution \(\underline{1.29}\) at. Vector \(\overrightarrow{\mathrm{A}}\) has a magnitude of 12.3 units and points due west. Vector \(\vec{B}\) points due north. (a) What is the magnitude of \(\overrightarrow{\mathbf{B}}\) if \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) has a magnitude of 15.0 units? (b) What is the direction of \(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}\) relative to due west? (c) What is the magnitude of \(\overrightarrow{\mathrm{B}}\) if \(\overrightarrow{\mathrm{A}} \overrightarrow{\mathrm{B}}\) has a magnitude of 15.0 units? (d) What is the direction of \(\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}\) relative to due west?

An observer, whose eyes are \(1.83 \mathrm{~m}\) above the ground, is standing \(32.0 \mathrm{~m}\) away from a tree. The ground is level, and the tree is growing perpendicular to it. The observer's line of sight with the treetop makes an angle of \(20.0^{\circ}\) above the horizontal. How tall is the tree?

The components of vector \(\overrightarrow{\mathbf{A}}\) are \(A_{x}\) and \(A_{v}\) (both positive), and the angle that it makes with respect to the positive \(x\) axis is \(\theta\). (a) Does increasing the component \(A_{x}\) (while holding \(A_{y}\) constant) increase or decrease the angle \(\theta\) ? (b) Does increasing the component \(A_{y}\) (while holding \(A_{x}\) constant) increase or decrease the angle \(\theta\) ? Account for your answers. The components of displacement vector \(\overrightarrow{\mathrm{A}}\) are \(A_{x}=12 \mathrm{~m}\) and \(A_{y}=12 \mathrm{~m}\). Find \(\theta\). (b) The components of displacement vector \(\overrightarrow{\mathrm{A}}\) are \(A_{x}=17 \mathrm{~m}\) and \(A_{y}=12 \mathrm{~m}\). Find \(\theta\). (c) The components of displacement vector \(\overrightarrow{\mathbf{A}}\) are \(A_{x}=12 \mathrm{~m}\) and \(A_{y}=17 \mathrm{~m}\). Find \(\theta\). Be sure that your answers are consistent with your answers to the Concept

Vector \(\overrightarrow{\mathrm{A}}\) points along the \(+y\) axis and has a magnitude of \(100.0\) units. Vector \(\overrightarrow{\mathbf{B}}\) points at an angle of \(60.0^{\circ}\) above the \(+x\) axis and has a magnitude of \(200.0\) units. Vector \(\vec{C}\) points along the \(+x\) axis and has a magnitude of \(150.0\) units. Which vector has (a) the largest \(x\) component and (b) the largest \(y\) component?

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