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Chapter 5: Problem 3

A body has an acceleration of \(3.00 \mathrm{~m} / \mathrm{s}^{2}\) at \(30.0^{\circ}\) to the positive direction of an \(x\) axis. The mass of the body is \(2.00 \mathrm{~kg}\). Find (a) the \(x\) component and (b) the \(y\) component of the net force acting on the body. (c) What is the net force in unit-vector notation?

Short Answer

Expert verified
(a) \(5.20 \text{ N}\), (b) \(3.00 \text{ N}\), (c) \(\mathbf{F} = 5.20 \mathbf{i} + 3.00 \mathbf{j} \text{ N}\)."

Step by step solution

01

Understand the Given Information

We are given the acceleration magnitude as \(3.00 \text{ m/s}^2\) at an angle of \(30.0^\circ\) relative to the positive \(x\) axis. The mass of the body is \(2.00 \text{ kg}\). We need to find the force components using Newton's second law \(F = ma\).
02

Calculate the Net Force

First, let's calculate the net force vector using the formula \(F = ma\), where \(m = 2.00 \text{ kg}\) and \(a = 3.00 \text{ m/s}^2\). Therefore, the magnitude of the net force is \( F = 2.00 \times 3.00 = 6.00 \text{ N}\).
03

Decompose Force into Components

Since the force is at an angle, we break it down using trigonometry. The \(x\) component is \( F_x = F \cos \theta \) and the \(y\) component is \( F_y = F \sin \theta \), where \(\theta = 30.0^\circ\).
04

Calculate the X Component of Force

Use \( F_x = F \cos \theta = 6.00 \cos 30.0^\circ\). Evaluating gives \( F_x = 6.00 \times \frac{\sqrt{3}}{2} \approx 5.20 \text{ N}\).
05

Calculate the Y Component of Force

Use \( F_y = F \sin \theta = 6.00 \sin 30.0^\circ\). Evaluating gives \( F_y = 6.00 \times 0.5 = 3.00 \text{ N}\).
06

Express the Force in Unit-Vector Notation

Combine the components with unit vectors: \(\mathbf{F} = F_x \mathbf{i} + F_y \mathbf{j} = 5.20 \mathbf{i} + 3.00 \mathbf{j} \text{ N}\).

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Key Concepts

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

Force Decomposition
When a force acts at an angle, it can be helpful to break it down into components. This simplifies the process of understanding the direction and magnitude of each part of the force. In physics, this technique is known as force decomposition. The reason we split a force into components is due to its vector nature. - Vectors have both magnitude and direction. - By breaking a force vector into components, we can easily analyze its effects in each direction separately. To decompose a force lying on a plane, we often use trigonometric functions. Here, sine and cosine are utilized to find the force's precise contributions along the x and y axes. This breakdown is crucial in solving problems where the forces are not aligned with the coordinate axes. Understanding how to decompose forces lays a solid foundation for tackling complex physics problems.
Vector Components
Vector components are the projections of a vector along the axes of a coordinate system. When you decompose a force, you are essentially determining its vector components. These components tell us how much of the force is acting along each direction, in this case, the x and y axes.- The x-component of the force can be found using the formula: \[ F_x = F \cos \theta \] Here, \(F\) is the magnitude of the force and \(\theta\) is the angle with the x-axis.- The y-component is determined by: \[ F_y = F \sin \theta \]This method leverages trigonometric ratios to separate the effects of the force in each axis direction, given its presence at an angle. By computing these components, you can apply Newton's second law, \( F = ma \), effectively along each axis, making problem-solving streamlined and intuitive.
Unit-Vector Notation
In physics, it鈥檚 not only important to quantify force components but also to express them clearly in vector form. Unit-vector notation is a way to express vectors using the standard unit vectors along the coordinate axes.- Unit vectors are denoted as \(\mathbf{i}\) for the x-axis and \(\mathbf{j}\) for the y-axis. They are vectors of unit length pointing in the direction of the respective axes.Expressing a force in unit-vector notation follows the formula:\[ \mathbf{F} = F_x \mathbf{i} + F_y \mathbf{j} \]This notation clearly spells out the contributions of the force along each axis, making it highly useful for both calculation and conceptual understanding. Unit-vector notation provides a standardized and clear way to communicate the direction and magnitude of vectors, which is essential for working with vector quantities in any physical analysis.

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Most popular questions from this chapter

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 percentage is the launch speed decreased if the athlete increases the angle from \(30.00^{\circ}\) to \(42.00^{\circ}\) ?

A \(1.50 \mathrm{~kg}\) object is subjected to three forces that give it an acceleration \(\vec{a}=-\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}}\). If two of the three forces are \(\vec{F}_{1}=(30.0 \mathrm{~N}) \hat{\mathrm{i}}+(16.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=-(12.0 \mathrm{~N}) \overrightarrow{\mathrm{i}}+(8.00 \mathrm{~N}) \hat{\mathrm{j}}\), find the third force.

The tension at which a fishing line snaps is commonly called the line's "strength." What minimum strength is needed for a line that is to stop a salmon of weight \(90 \mathrm{~N}\) in \(11 \mathrm{~cm}\) if the fish is initially drifting at \(2.8 \mathrm{~m} / \mathrm{s}\) ? Assume a constant deceleration.

Figure \(5-35\) shows a \(5.00 \mathrm{~kg}\) block being pulled along a frictionless floor by a cord that applies a force of constant magnitude \(15.0 \mathrm{~N}\) but with an angle \(\theta(t)\) that varies with time. When angle \(\theta=\) \(25.0^{0}\), at what rate is the acceleration of the block changing if (a) \(\theta(t)=\left(2.00 \times 10^{-2} \mathrm{deg} / \mathrm{s}\right) t\) and \((\mathrm{b}) \theta(t)=-\left(2.00 \times 10^{-2} \mathrm{deg} / \mathrm{s}\right) t ?\) (Hint: The angle should be in radians)

A \(50 \mathrm{~kg}\) skier skis directly down a frictionless slope angled at \(10^{\circ}\) to the horizontal Assume the skier moves in the negative direction of an \(x\) axis along the slope. A wind force with component \(F_{x}\) acts on the skier. What is \(F_{x}\) if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\), and (c) increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) ?
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