Math Problem Statement

The 59.8cm diameter disk rotates counterclockwise on an axle through its center. F1=45.8N, F2=68.3N, F3=68.3N, F4=45.8N, and d=3.42cm. What is the net torque about the axle? Let the counterclockwise direction be positive.

Solution

The problem presents a disk rotating counterclockwise with four forces acting on it, and asks for the net torque about the axle. Let's solve this step by step:

Given Data:

  • Diameter of the disk = 59.8 cm, so the radius r=59.82=29.9r = \frac{59.8}{2} = 29.9 cm = 0.299 m
  • Distance from the center to the forces (d) = 3.42 cm = 0.0342 m
  • Forces:
    • F1=45.8NF_1 = 45.8 \, \text{N}
    • F2=68.3NF_2 = 68.3 \, \text{N}
    • F3=68.3NF_3 = 68.3 \, \text{N}
    • F4=45.8NF_4 = 45.8 \, \text{N}
  • Angle for F4F_4 is 4545^\circ
  • Counterclockwise torque is taken as positive.

Torque Formula:

The torque (τ\tau) due to a force is given by: τ=rFsin(θ)\tau = r F \sin(\theta) where rr is the perpendicular distance from the axis of rotation to the point of application of the force, FF is the force, and θ\theta is the angle between the force and the radius.

Torque due to each force:

  1. Force F1F_1:

    • Acts vertically downward (angle 270270^\circ relative to positive x-axis)
    • Torque τ1=rF1sin(90)=0.299×45.8=13.69Nm\tau_1 = -r F_1 \sin(90^\circ) = -0.299 \times 45.8 = -13.69 \, \text{N}\cdot\text{m}
  2. Force F2F_2:

    • Acts vertically upward (angle 9090^\circ relative to positive x-axis)
    • Torque τ2=rF2sin(90)=0.299×68.3=20.42Nm\tau_2 = r F_2 \sin(90^\circ) = 0.299 \times 68.3 = 20.42 \, \text{N}\cdot\text{m}
  3. Force F3F_3:

    • Acts horizontally to the left (angle 180180^\circ)
    • No torque contribution because the force is directed along the axis (sin(180)=0\sin(180^\circ) = 0).
  4. Force F4F_4:

    • Acts at 4545^\circ from the x-axis
    • Torque τ4=rF4sin(45)=0.299×45.8×sin(45)=9.68Nm\tau_4 = r F_4 \sin(45^\circ) = 0.299 \times 45.8 \times \sin(45^\circ) = 9.68 \, \text{N}\cdot\text{m}

Net Torque:

Now, summing up all the torques: τnet=τ1+τ2+τ3+τ4\tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 + \tau_4 Since τ3=0\tau_3 = 0, we have: τnet=13.69+20.42+0+9.68=16.41Nm\tau_{\text{net}} = -13.69 + 20.42 + 0 + 9.68 = 16.41 \, \text{N}\cdot\text{m} Thus, the net torque about the axle is 16.41Nm16.41 \, \text{N}\cdot\text{m} in the counterclockwise direction.

Would you like more details on this solution, or do you have any questions? Here are some related questions to expand on this:

  1. What is the significance of the direction of the torque in physical systems?
  2. How would the solution change if the disk rotated clockwise instead of counterclockwise?
  3. What would happen if the forces were not perpendicular to the radius?
  4. How is torque related to angular acceleration in rotational motion?
  5. What other physical factors, such as friction, could affect the net torque in real-life situations?

Tip: Always ensure to calculate the correct direction of each torque when dealing with multiple forces, as this can affect the overall result!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Torque in rotational motion
Vectors and angles in physics

Formulas

τ = r × F × sin(θ)

Theorems

Torque formula in rotational dynamics
Trigonometric function for angle of force application

Suitable Grade Level

College/University level (Physics)