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Chapter 57: Practice Exercises

- A girl is sitting on a seesaw near the balance point.

- The stick is balanced.
- The 10-g mass is shown in the distance from the fulcrum.

- A solid cylinder consisting of an outer and an inner radius r 1 and r 2 is pivoted on a axle.

- Answer the questions based on the diagram.
- Massless is what the rod is considered to be.

- A net Torque of T is experienced by an object with a moment of inertia.

- A person is 2.5 m from a wall with a beam attached.
- The beam weighs 200 N and is 6 m long.
- A cable that is attached to the free end of the beam is attached to the wall.

- A ladder of length l and weight 100 N rests against a wall.
- The floor and the bottom of the ladder have the same coefficients of static friction.

- A toy top is rotating at a rate of 8 times per second and has a friction Torque of 0.2.

- The force on each side is determined by the weight.
- The arm distances are measured in m and x.

- The same factors appear on both sides of the balance equation, but we could convert all of the mass to kilograms and all of the distance to meters.

- F-2 has a counterclockwise positive Torque and F-1 has a clockwise negative Torque.
- The necessary arm distance is determined by each radius.

- The components of the remaining forces to the beam are needed.
- The 30-N force acts counterclockwise, while the 10-N force acts clockwise.

- The 10-N force will act in a clockwise direction, while the 20-N force will act in a counterclockwise direction.
- The pivot is 1.5 m from each force.
- The component of each force has to be parallel to the beam.

- We get T in the last equation.

- The ratio of these two equations gives us a tan of 1.27 and a th of 51.8deg.

- The force P is the normal force of the wall.
- The weight Fg acts from the center of the mass.
- We can write equations for the conditions of static equilibrium since we have equilibrium at this angle.

- The sum of both x and y must be zero.
- Friction is an opposing force and the reaction force.
- We can write 0 if we want to.
- Since Fg is 100 N, then N is 100 N.

- The sum of the Torques must not be more than zero.
- The component of P needs to be parallel to the ladder.
- We can see from the geometry that this is P sinth.
- The component of weight on the ladder is Fg cos th.

- The length of the ladder is irrelevant and can be canceled out.
- We know that P is 50 N and Fg is 100 N.

- An object has stopped rotating when it is not moving.

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