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Butoxors [25]
3 years ago
15

A bicycle pedal is pushed straight downwards by a foot with a 11 Newton force. The shaft of the pedal is 20 cm long. If the shaf

t is π/6 radians past horizontal, what is the magnitude of the torque about the point where the shaft is attached to the bicycle?
Mathematics
1 answer:
galben [10]3 years ago
3 0

Answer: 1.1 Nm

Step-by-step explanation:

The formula for torque is T = r · F · sin θ

where θ = π/6, r = 20 cm = 0.2 m, F = 11 N

formula for converting radians to degrees is 1 rad × 180/π = 57.296°

θ = π/6 · 180/π = 30°

Substituting r into the formula for torque, we have

T = 0.2 x 11 x sin 30° = 0.2 x 11 x 0.5

T = 1.1 Nm

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General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

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Step-by-step explanation:

<u>Step 1: Define</u>

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<u>Step 2: Find </u><em><u>r</u></em>

  1. Substitute [SAS]:                    23 in² = 4πr²
  2. Isolate <em>r </em>term:                         23 in²/(4π) = r²
  3. Isolate <em>r</em>:                                  √[23 in²/(4π)] = r
  4. Rewrite:                                   r = √[23 in²/(4π)]
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