Math Problem Statement

if a 10 kg weight has 10 newtons applied at 45 degrees, how to get its trajectory?

Solution

To determine the trajectory of a 10 kg weight with a 10 N force applied at a 45-degree angle, we need to follow a step-by-step process rooted in Newton's laws of motion and kinematic equations. Here's how to approach it:

Step 1: Resolve the force into components

The force applied at an angle can be broken down into horizontal (FxF_x) and vertical (FyF_y) components using trigonometry.

  • Fx=Fcos(θ)=10Ncos(45)=10N227.07NF_x = F \cdot \cos(\theta) = 10 \, \text{N} \cdot \cos(45^\circ) = 10 \, \text{N} \cdot \frac{\sqrt{2}}{2} \approx 7.07 \, \text{N}
  • Fy=Fsin(θ)=10Nsin(45)=10N227.07NF_y = F \cdot \sin(\theta) = 10 \, \text{N} \cdot \sin(45^\circ) = 10 \, \text{N} \cdot \frac{\sqrt{2}}{2} \approx 7.07 \, \text{N}

Step 2: Calculate the acceleration

Next, use Newton's second law to find the acceleration components.

  • Horizontal acceleration: ax=Fxm=7.07N10kg=0.707m/s2a_x = \frac{F_x}{m} = \frac{7.07 \, \text{N}}{10 \, \text{kg}} = 0.707 \, \text{m/s}^2
  • Vertical acceleration: ay=Fymg=7.07N10kg9.8m/s2=9.093m/s2a_y = \frac{F_y}{m} - g = \frac{7.07 \, \text{N}}{10 \, \text{kg}} - 9.8 \, \text{m/s}^2 = -9.093 \, \text{m/s}^2 (considering gravity acting downward with g=9.8m/s2g = 9.8 \, \text{m/s}^2)

Step 3: Use kinematic equations

To calculate the trajectory, we use the following kinematic equations for projectile motion:

  1. Horizontal motion (no horizontal acceleration besides axa_x):

    • x(t)=v0xt+12axt2x(t) = v_{0x} t + \frac{1}{2} a_x t^2
  2. Vertical motion (including gravitational acceleration):

    • y(t)=v0yt+12ayt2y(t) = v_{0y} t + \frac{1}{2} a_y t^2

For simplicity, assume initial velocities v0x=v0y=0v_{0x} = v_{0y} = 0 (assuming the object is at rest before the force is applied). So the equations simplify to:

  • x(t)=12axt2=12(0.707)t2x(t) = \frac{1}{2} a_x t^2 = \frac{1}{2} (0.707) t^2
  • y(t)=12ayt2=12(9.093)t2y(t) = \frac{1}{2} a_y t^2 = \frac{1}{2} (-9.093) t^2

Step 4: Time of flight and trajectory shape

To find the full trajectory, you need to determine how long the object remains in motion (time of flight) and the horizontal distance it covers. You can solve the above equations for different values of time tt to get the position coordinates (x,y)(x, y).

For example, solving for y(t)=0y(t) = 0 will give you the time when the object hits the ground again, and you can substitute that time into x(t)x(t) to find the horizontal distance.

Do you want a step-by-step breakdown with numerical integration, or shall I proceed further on specific aspects of this problem?


Questions to further explore this topic:

  1. How does gravity affect the vertical motion of the object compared to the horizontal motion?
  2. What happens to the trajectory if the angle of the applied force changes?
  3. How would air resistance alter the trajectory?
  4. Can we use energy conservation to predict any part of the object's motion?
  5. What are the differences in trajectory if the mass of the object changes but the force remains constant?

Tip: Always resolve forces into components when dealing with vectors at an angle.

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Math Problem Analysis

Mathematical Concepts

Vector Decomposition
Newton's Laws
Projectile Motion
Trigonometry

Formulas

F_x = F * cos(θ)
F_y = F * sin(θ)
a_x = F_x / m
a_y = F_y / m - g
x(t) = v_{0x} t + (1/2) a_x t^2
y(t) = v_{0y} t + (1/2) a_y t^2

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 10-12