Problem 29 - College Physics

Question

A 2.0 -kg object has an acceleration of \(1.5 \mathrm{~m} / \mathrm{s}^{2}\) at \(30^{\circ}\) above the \(-x\) -axis. Write the force vector producing this acceleration in component form.

Step-by-Step Solution

Answer

The force vector is \((-2.598, 1.5)\) N.

Step 1: Understand the problem

We need to find the force vector in component form that causes a 2.0 kg object to accelerate at 1.5 m/s² at an angle of 30° above the negative x-axis.

Step 2: Use Newton's Second Law

Newton's Second Law states that force is mass times acceleration. The formula is given by \( F = m imes a \), where \( F \) is the force vector, \( m \) is the mass, and \( a \) is the acceleration vector.

Step 3: Calculate the magnitude of the force

The magnitude of the force can be calculated by using \( F = m imes a \). For this problem, \( m = 2.0 \text{ kg} \) and \( a = 1.5 \text{ m/s}^2 \). Thus, the magnitude of the force is \( F = 2.0 imes 1.5 = 3.0 \text{ N} \).

Step 4: Find the force components

We have the force vector's magnitude and angle. To find the components, use sine and cosine:- The x-component (opposite -x axis) is \( F_x = F imes ext{cos}(30°) = -3.0 \times rac{ ext{cos}(30°)}{1} \).- The y-component is \( F_y = F imes ext{sin}(30°) = 3.0 \times ext{sin}(30°) \).

Step 5: Calculate the x-component

Evaluate \( F_x = 3.0 \times ext{cos}(30°) \). Since cos(30°) = \( rac{ ext{sqrt}(3)}{2} \),\( F_x = -3.0 \times rac{ ext{sqrt}(3)}{2} = -rac{3.0 ext{sqrt}(3)}{2} \approx -2.598 \text{ N} \).

Step 6: Calculate the y-component

Evaluate \( F_y = 3.0 \times ext{sin}(30°) \). Since sin(30°) = \( rac{1}{2} \),\( F_y = 3.0 \times rac{1}{2} = 1.5 \text{ N} \).

Step 7: Write the force vector in component form

The force vector \( \mathbf{F} \) is given by \( \mathbf{F} = (F_x, F_y) = (-2.598, 1.5) \text{ N} \).

Key Concepts

Force Vector Components

Vectors have both magnitude and direction, making them essential in physics for understanding forces. When dealing with forces, particularly in two dimensions, it's often necessary to break the force vector into its components along the x and y axes. This process simplifies the calculations and helps in understanding how the force affects motion in each direction separately.

To determine the force vector components, first calculate the magnitude of the overall force using Newton's Second Law. Once you have the magnitude, use trigonometry combined with an angle to find the horizontal (x-axis) and vertical (y-axis) components:

• The x-component: This is given by the product of the force's magnitude and the cosine of the angle the force makes with the reference axis (In our exercise, this angle is measured from the negative x-axis).
• The y-component: This comes from multiplying the force's magnitude by the sine of the same angle.

The force vector is therefore expressed in component form as \(\mathbf{F} = (F_x, F_y)\). This form makes it straightforward to analyze the force's effect on the object in each independent direction.

Trigonometry in Physics

Trigonometry is a powerful tool used in physics to resolve forces into their components, especially when they act at angles. Let's explore how this applies to our exercise:

When a force acts at an angle, such as 30° above the negative x-axis, the force vector can be split into two perpendicular components using trigonometric functions:

• Sine function: In this context, it helps determine the y-component of the force, by multiplying the sine of the angle with the magnitude of the force.
• Cosine function: It assists in calculating the x-component by using the cosine of the angle, again multiplied by the force magnitude.

In our example, we position the angle within a coordinate system where the cosine function identifies the direction along the x-axis and the sine offers the y-axis direction. Mastery of these functions is crucial as it allows for determining how exactly a force influences an object's movement, whether upward, downward, or in a horizontal plane.

Acceleration and Force Calculation

Understanding how to calculate acceleration and force in physics begins with grasping Newton's Second Law, which connects these concepts. This law can be eloquently formulated as \( F = m \times a \), where:\

• \(F\) is the force applied to an object, measured in Newtons (N).
• \(m\) is the mass of the object in kilograms (kg).
• \(a\) is the acceleration, measured in meters per second squared (m/s²).

For our specific problem, we knew the mass (2.0 kg) and the acceleration (1.5 m/s²). Multiplying them gives us the magnitude of the force. In situations where the acceleration isn't aligned with the axes, further trigonometry, as covered previously, distributes this force into its effective components.

Remember, Newton's Second Law doesn't just calculate force; it's integral in predicting how forces affect the movement of an object relative to its mass. This understanding is foundational in dynamics and helps clarify how objects interact and move under various forces.