The length of a bicycle pedal arm is 0.152 m, and a downward force of 146 N is applied to the pedal by the rider's foot. What is the magnitude of the torque about the pedal arm's pivot point when the arm makes the following angles with the vertical?
(a) 30°
(b) 90°
(c) 180°
And what is your thinking. Providing answers to questions is not in my opinion very helpful to you in the long run. I will gladly critique your thinking. Thinking is NOT "I think the answer is ..." unless you say " because..."
Thanks.
I figured it out. T=r*F*sin theta. Thanks. Please help me out on my problem posted above.
To find the magnitude of the torque about the pedal arm's pivot point, you can use the formula T = r * F * sin(θ), where T is the torque, r is the length of the pedal arm, F is the force applied, and θ is the angle between the arm and the vertical.
Now, let's calculate the torque for each angle given:
(a) When θ = 30°:
T = (0.152 m) * (146 N) * sin(30°)
T = 0.152 m * 146 N * 0.5
T ≈ 11.16 N·m
(b) When θ = 90°:
T = (0.152 m) * (146 N) * sin(90°)
T = 0.152 m * 146 N * 1
T ≈ 22.192 N·m
(c) When θ = 180°:
T = (0.152 m) * (146 N) * sin(180°)
T = 0.152 m * 146 N * 0
T = 0 N·m
Therefore, the magnitude of the torque for each angle is:
(a) 11.16 N·m
(b) 22.192 N·m
(c) 0 N·m
I hope this helps! Let me know if you have any further questions.