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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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