What is the angular speed of the tip of the minute hand on a clock, in rad/s?
1 answer:
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Refer to the diagram shown below.
The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N
Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m
This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
F = 4.04 N
Answer: 4.04 N
The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N
Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m
This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
F = 4.04 N
Answer: 4.04 N