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A bucket filled with water has a mass of 43 kg and is hanging from a rope that is wound around a stationary cylinder of radius 0.043 m. If the cylinder does not rotate and the bucket hangs straight down, what is the magnitude of the torque produced by the bucket around the center of the cylinder? The acceleration of gravity is 9.81 m/s 2 . Answer in units of N · m

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cjrodriguez89

Answer:

18.1 N*m

Explanation:

  • The torque exerted by a force, regarding a given point, in magnitude, is given by the following expression:

        [tex]\tau = F*r*sin \theta (1)[/tex]

  • where r is the distance between the point of application of the force to the point where the torque must be calculated, and θ is the angle between the direction of the force and the vector r, representing this distance.
  • The force F is just the tension on the rope, and is equal to the weight of the bucket, as follows:

       [tex]F = m_{b} * g = 43 kg * 9.81 m/s2 = 421. 8 N[/tex]

  • In this case, r is simply the radius of the cylinder:

        r = 0.043 m

  • As the bucket is hanging straight down, the tension on the rope is perfectly vertical, so the radius from the center of the cylinder to the point of application of the force, is horizontal.
  • So, r and F and perpendicular each other so, sin θ =1
  • Replacing by the givens in (1), we can solve for τ, as follows:

        [tex]\tau = m*g*r = 43 kg* 9.81 m/s2* 0.043 m\\ \\ \tau = 18. 1 N*m[/tex]

  • The magnitude of the torque produced by the bucket is 18.1 N*m.

Answer:

18.14 Nm

Explanation:

Torque is the product of force applied and the perpendicular distance from the pivot (point of rotation) to point of application of force.

T = F*d

here the force is the weight which acts downwards perpendicular to the horizontal distance from the center of rotation or cylinder

F = weight = mg = 43 * 9.81 = 421.83 N

d = moment arm = radius of cylinder = 0.043 m

Torque = T = F*d = 421.83 * 0.043 = 18.138 Nm

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