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Answer :
Final answer:
The torque exerted by a 6.60 kg banner hanging from a 1.80 m long horizontal pole is calculated using the formula for torque with gravity acting directly downward, resulting in an answer of 116 Nm.
Hence, the correct answer is option 2.
Explanation:
The question is asking for the torque exerted by the weight of a banner on a horizontal pole. To calculate the torque (T), we use the formula T = rF sin θ, where r is the lever arm distance, F is the force, and θ is the angle between the force and the lever arm. Here, the weight of the banner is the force, which can be calculated as F = m × g (mass times the acceleration due to gravity).
In this case, the lever arm is the length of the pole (1.80 m), the mass of the banner (6.60 kg) is the weight at the end of the pole, and the angle θ is 90 degrees since the force due to gravity acts directly downward. As sin 90° is 1, the formula simplifies to T = r × m × g. By substituting the given values and using g = 9.8 m/s², we calculate the torque exerted by the weight of the banner as follows: T = 1.80 m × 6.60 kg × 9.8 m/s² = 116.424 N·m. Hence, the correct answer is 116 Nm.
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