Math Problem Statement

In the image shown, the right-hand 87 N force is applied perfectly vertical (+y) at a distance d1 = 0.32 m from the axis of rotation of the wrench (red dot). If the same force were to be applied at a distance d2 = 0.28 m from the axis instead (the left-hand force as shown), at what angle θ would this force need to be applied to produce the same torque or moment about the axis of rotation?

Solution

The problem you're dealing with involves torque, which is the rotational equivalent of force. The torque (τ\tau) is given by the equation:

τ=Fdsin(θ)\tau = F \cdot d \cdot \sin(\theta)

Where:

  • FF is the force applied,
  • dd is the distance from the axis of rotation,
  • θ\theta is the angle between the force direction and the lever arm,
  • τ\tau is the torque.

Given:

  • Force F=87NF = 87 \, \text{N},
  • Initial distance d1=0.32md_1 = 0.32 \, \text{m} with the force applied vertically (θ=90\theta = 90^\circ) meaning sin(θ)=1\sin(\theta) = 1,
  • New distance d2=0.28md_2 = 0.28 \, \text{m}, and the goal is to find the angle θ\theta for the same torque.

Step 1: Calculate the initial torque

For the vertical force: τ1=Fd1sin(90)=87N0.32m1=27.84Nm\tau_1 = F \cdot d_1 \cdot \sin(90^\circ) = 87 \, \text{N} \cdot 0.32 \, \text{m} \cdot 1 = 27.84 \, \text{Nm}

Step 2: Calculate the angle θ\theta

For the second case, the same torque τ2=τ1\tau_2 = \tau_1 needs to be generated, but with the force at an unknown angle θ\theta:

τ2=Fd2sin(θ)\tau_2 = F \cdot d_2 \cdot \sin(\theta) 27.84=870.28sin(θ)27.84 = 87 \cdot 0.28 \cdot \sin(\theta)

Now, solve for sin(θ)\sin(\theta):

sin(θ)=27.84870.28=27.8424.361.142\sin(\theta) = \frac{27.84}{87 \cdot 0.28} = \frac{27.84}{24.36} \approx 1.142

Since sin(θ)\sin(\theta) cannot exceed 1, this implies an error or oversimplification. Let's check if there's an assumption we missed.

Would you like a further breakdown of any steps or refinements on the details of this specific setup?


Related Questions:

  1. How does torque change with distance from the axis of rotation?
  2. Why can sin(θ)\sin(\theta) not exceed 1, and what physical meaning does that have here?
  3. What would happen if the force was applied at a different point along the wrench?
  4. How would the result change if the force was not applied vertically but at an arbitrary angle initially?
  5. Could a smaller force applied farther from the axis achieve the same torque?

Tip: In torque calculations, make sure to check if the system's constraints allow all trigonometric functions to stay within their defined ranges.

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Math Problem Analysis

Mathematical Concepts

Torque
Trigonometry

Formulas

Torque formula: τ = F ⋅ d ⋅ sin(θ)
Sine function properties

Theorems

Trigonometric functions and ranges

Suitable Grade Level

Grades 11-12 or introductory university physics