/*! 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 4 A force of \(100 \mathrm{~N}\) m... [FREE SOLUTION] | 91Ó°ÊÓ

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A force of \(100 \mathrm{~N}\) makes an angle of \(\theta\) with the \(x\) -axis and has a scalar \(y\) -component of \(30 \mathrm{~N}\). Find both the scalar \(x\) -component of the force and the angle \(\theta\). (Remember that the number \(100 \mathrm{~N}\) has three significant figures whereas \(30 \mathrm{~N}\) has only two.)Begin your analysis by drawing a diagram. Here the data are sketched roughly in Fig. \(3-4\). We wish to find \(F_{x}\) and \(\theta\). Since $$ \sin \theta=\frac{30 \mathrm{~N}}{100 \mathrm{~N}}=0.30 $$ \(\theta=17.46^{\circ}\), and thus, to two significant figures, \(\theta=17^{\circ} .\) Then, using the \(\cos \theta\) $$ F_{x}=(100 \mathrm{~N}) \cos 17.46^{\circ}=95 \mathrm{~N} $$

Short Answer

Expert verified
The x-component is 95 N and the angle \( \theta \) is 17°.

Step by step solution

01

Identify the variables and known information

We are given a force of magnitude 100 N that makes an angle \( \theta \) with the x-axis. The y-component of the force is 30 N. We are to find both the x-component \( F_x \) and the angle \( \theta \).
02

Calculate the angle using sine function

We use the sine function to find the angle \( \theta \), where \( \sin \theta = \frac{30 \text{ N}}{100 \text{ N}} = 0.30 \). This gives us \( \theta = \sin^{-1}(0.30) \), which calculates to \( \theta = 17.46^{\circ} \). Rounding to two significant figures, \( \theta = 17^{\circ} \).
03

Calculate the x-component using cosine function

The x-component \( F_x \) can be calculated using the cosine function: \( F_x = (100 \text{ N}) \cos 17.46^{\circ} \). Using a calculator, this evaluates to \( F_x \approx 95 \text{ N} \).

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Key Concepts

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

Understanding Trigonometry in Vector Decomposition
Trigonometry plays a crucial role in vector decomposition, particularly when you are dealing with force vectors in physics. When a force is applied at an angle, it is often useful to decompose it into its horizontal (x) and vertical (y) components. This is because analyzing these components separately can greatly simplify complex problems.

To find these components, you need to use basic trigonometric functions such as sine, cosine, and tangent. In this scenario, we have a known total force and a y-component. We use the sine function to determine the angle of the force:
  • Sine (\(\sin\theta\)) relates the opposite side (y-component) to the hypotenuse (total force).
  • Cosine (\(\cos\theta\)) relates the adjacent side (x-component) to the hypotenuse.
To find the angle \(\theta\), you calculate \(\sin\theta = \frac{30\,\text{N}}{100\,\text{N}}\). The inverse sine function \(\sin^{-1}(0.30)\) then gives you the angle \(\theta = 17.46^{\circ}\). This is a simple yet powerful application of trigonometry in physics problem solving.

Further decomposition using cosine lets us find the x-component: \(\cos\theta = \frac{F_x}{100\,\text{N}}\), leading to \(F_x = (100\,\text{N}) \cos 17.46^{\circ} \approx 95\,\text{N}\). Understanding how to effectively apply these trigonometric functions will enhance your problem-solving skills in physics.
Significance of Significant Figures in Calculations
Significant figures are essential in ensuring that your calculations reflect the precision of the data you have. Knowing how to handle significant figures will help you communicate the accuracy of your results correctly.

When working with measurements like forces, it's typical to encounter different levels of precision. In our problem:
  • The total force is given as 100 N, with three significant figures.
  • The y-component is 30 N, with only two significant figures.
The precision of your calculated results should not exceed that of your least precise measurement. When calculating \(\theta\), it is initially computed as \(17.46^{\circ}\). However, because 30 N is to two significant figures, we round the angle to \(17^{\circ}\).

Similarly, while the x-component is calculated as \(95.0\,\text{N}\) from a calculation, we should also consider that the original measurement allowed only for two significant figures, so it remains \(95\,\text{N}\) in full context. By adhering to these rules, you ensure that your results maintain their scientific credibility.
Steps for Effective Physics Problem Solving
Solving physics problems efficiently requires a structured approach. This ensures that you tackle each problem methodically and accurately. Below are some steps that exemplify a systematic approach, as demonstrated in our vector decomposition problem:

Start by breaking down the problem's requirements. Identify what is given and what needs to be found. Here, we knew the total force and y-component and needed both the x-component and angle of the force.

Next, apply relevant formulas or principles. For this problem, we used trigonometric relationships—\(\sin\) and \(\cos\)—to determine the missing components. This involved calculating the angle using \(\sin\) first, followed by obtaining the x-component through \(\cos\).

Once you have your calculations, check if they make sense. Verifying units and ensuring the right number of significant figures can prevent errors. It’s also helpful to reason through the problem: Does the angle seem reasonable for the given y-component? Is the x-component consistent with the force's direction?

Lastly, review and reflect on what you've done, aiming to understand each step's role in solving the problem. This reflective practice helps solidify your understanding and prepares you for tackling similar problems in the future.

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Most popular questions from this chapter

An inclined plane makes an angle of \(30^{\circ}\) with the horizontal. Find the constant force, applied parallel to the plane, required to cause a \(15-\mathrm{kg}\) box to slide \((a)\) up the plane with acceleration \(1.2 \mathrm{~m} / \mathrm{s}^{2}\) and \((b)\) down the incline with acceleration \(1.2 \mathrm{~m} / \mathrm{s}^{2} .\) Neglect friction forces.

A constant force acts on a \(5.0 \mathrm{~kg}\) object and reduces its velocity from \(7.0 \mathrm{~m} / \mathrm{s}\) to \(3.0 \mathrm{~m} / \mathrm{s}\) in a time of \(3.0 \mathrm{~s}\). Determine the force. We must first find the acceleration of the object, which is constant because the force is constant. Taking the direction of motion as positive, from Chapter 2 $$ a=\frac{v_{f}-v_{i}}{t}=\frac{-4.0 \mathrm{~m} / \mathrm{s}}{3.0 \mathrm{~s}}=-1.33 \mathrm{~m} / \mathrm{s}^{2} $$ Use \(F=m a\) with \(m=5.0 \mathrm{~kg}\) : $$ F=(5.0 \mathrm{~kg})\left(-1.33 \mathrm{~m} / \mathrm{s}^{2}\right)=-6.7 \mathrm{~N} $$ The minus sign indicates that the force is a retarding force, directed opposite to the motion.

Find the weight on the surface of the Earth of a body whose mass is (a) \(3.00 \mathrm{~kg}\), and \((b) 200 \mathrm{~g}\). The general relation between mass \(m\) and weight \(F_{W}\) is \(F_{W}=m g\). In this relation, \(m\) must be in kilograms, \(g\) in meters per second squared, and \(F_{W}\) in newtons. On Earth, \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). The acceleration due to gravity varies from place to place in the universe. (a) \(\quad F_{W}=(3.00 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=29.4 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}^{2}=29.4 \mathrm{~N}\) (b) \(F_{W}=(0.200 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=1.96 \mathrm{~N}\)

Compute the mass of the Earth, assuming it to be a sphere of radius \(6370 \mathrm{~km}\). Use the fact that \(\mathrm{g}=9.81 \mathrm{~m} / \mathrm{s}^{2}\) and give your answer to three significant figures. Let \(M_{E}\) be the mass of the Earth, and \(m\) the mass of an object. The weight of the object on the planet's surface is equal to \(m g .\) It is also equal to the gravitational force \(G\left(M_{E} m\right) R_{E}^{2}\), where \(R_{E}\) is the Earth's radius. Hence, $$ m g=G \frac{M_{E} m}{R_{E}^{2}} $$ from which \(\quad M_{E}=\frac{g R_{E}^{2}}{G}=\frac{\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\left(6.37 \times 10^{6} \mathrm{~m}\right)^{2}}{6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}}=5.97 \times 10^{24} \mathrm{~kg}\)

Compute algebraically the resultant of the following coplanar forces: \(100 \mathrm{~N}\) at \(30^{\circ}, 141.4 \mathrm{~N}\) at \(45^{\circ}\), and \(100 \mathrm{~N}\) at \(240^{\circ}\). Check your result graphically.

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