Dynamics - Polson 7-8
... • A man with a mass of 74 kg slides down a metal pole. If his acceleration is 0.38 m/s2 downward, what is the magnitude of the upward force exerted by friction? Gravity is the only other force to consider. ...
... • A man with a mass of 74 kg slides down a metal pole. If his acceleration is 0.38 m/s2 downward, what is the magnitude of the upward force exerted by friction? Gravity is the only other force to consider. ...
Part II
... look as if the reference frame is an inertial one, it’s necessary to introduce Fictitious Forces. – Technically, it’s the coordinate transformation from the inertial frame to the non-inertial one introduces terms on the “ma” side of ∑F = mainertial. If we want eqtns in the non-inertial frame to look ...
... look as if the reference frame is an inertial one, it’s necessary to introduce Fictitious Forces. – Technically, it’s the coordinate transformation from the inertial frame to the non-inertial one introduces terms on the “ma” side of ∑F = mainertial. If we want eqtns in the non-inertial frame to look ...
Transparancies for Dynamics - University of Manchester
... e.g. Alice walks forwards along a boat at 1m/s and the boat moves at 2 m/s. what is Alices’ velocity as seen by Bob ? • If Bob is on the boat it is just 1 m/s • If Bob is on the shore it is 1+2=3m/s • If Bob is on a boat passing in the opposite direction….. and the earth is spinning… ...
... e.g. Alice walks forwards along a boat at 1m/s and the boat moves at 2 m/s. what is Alices’ velocity as seen by Bob ? • If Bob is on the boat it is just 1 m/s • If Bob is on the shore it is 1+2=3m/s • If Bob is on a boat passing in the opposite direction….. and the earth is spinning… ...
amanda`sNewton`s First Law
... Sir Isaac Newton described the relationship between motion and force in the laws that we now call Newton’s laws of motion. His laws apply to a wide range of motion like a caterpillar crawling on a leaf, a person riding a bicycle, or a rocket blasting off into space. ...
... Sir Isaac Newton described the relationship between motion and force in the laws that we now call Newton’s laws of motion. His laws apply to a wide range of motion like a caterpillar crawling on a leaf, a person riding a bicycle, or a rocket blasting off into space. ...
NEWTON`S LAWS OF MOTION
... -Gravitational Force (or weight = mg where g is 9.8 m/s2) - “Normal forces” (one object touching another). 2. Draw a “Freebody Diagram” -draw the object, show all forces acting on that object as vectors pointing in the correct direction. Show the direction of the ...
... -Gravitational Force (or weight = mg where g is 9.8 m/s2) - “Normal forces” (one object touching another). 2. Draw a “Freebody Diagram” -draw the object, show all forces acting on that object as vectors pointing in the correct direction. Show the direction of the ...
NEWTON'S LAWS OF MOTION
... -Gravitational Force (or weight = mg where g is 9.8 m/s2) - “Normal forces” (one object touching another). 2. Draw a “Freebody Diagram” -draw the object, show all forces acting on that object as vectors pointing in the correct direction. Show the direction of the ...
... -Gravitational Force (or weight = mg where g is 9.8 m/s2) - “Normal forces” (one object touching another). 2. Draw a “Freebody Diagram” -draw the object, show all forces acting on that object as vectors pointing in the correct direction. Show the direction of the ...
Section 1 Forces Newton`s Second Law
... 1. A marble is placed at the top of a smooth ramp. What will happen to the marble? What force causes this? 2. A marble is rolling around in the back of a small toy wagon as the wagon is pulled along the sidewalk. When the wagon is stopped suddenly by a rock under one of the wheels, the marble rolls ...
... 1. A marble is placed at the top of a smooth ramp. What will happen to the marble? What force causes this? 2. A marble is rolling around in the back of a small toy wagon as the wagon is pulled along the sidewalk. When the wagon is stopped suddenly by a rock under one of the wheels, the marble rolls ...
Rotation Torque, Rolling, & Angular Momentum
... Two 2.00 kg balls are attached to the ends of a thin rod of length 50.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (see figure), a 50.0 g wad of wet putty drops onto one of the b ...
... Two 2.00 kg balls are attached to the ends of a thin rod of length 50.0 cm and negligible mass. The rod is free to rotate in a vertical plane without friction about a horizontal axis through its center. With the rod initially horizontal (see figure), a 50.0 g wad of wet putty drops onto one of the b ...
Application of Definite Integrals
... A force F = 3x2 - 2x acts on a body moving along the x-axis. Here x is in meters and F is in Newtons (N). If the force is applied at an angle of 300 with the direction of motion, find the work done by the force in moving the object from x = 1m to x = 2m. Solution: The work is calculated from equatio ...
... A force F = 3x2 - 2x acts on a body moving along the x-axis. Here x is in meters and F is in Newtons (N). If the force is applied at an angle of 300 with the direction of motion, find the work done by the force in moving the object from x = 1m to x = 2m. Solution: The work is calculated from equatio ...
Torque
... counterclockwise (CCW) depending which way the torque would make the object rotate about the axis of rotation chosen. Either CW or CCW can be chosen as the “+” direction as long as consistency is maintained. We will choose CW as the “+” direction here. The resultant of multiple torques can then be o ...
... counterclockwise (CCW) depending which way the torque would make the object rotate about the axis of rotation chosen. Either CW or CCW can be chosen as the “+” direction as long as consistency is maintained. We will choose CW as the “+” direction here. The resultant of multiple torques can then be o ...
CH4 Newton`s laws
... When a net external force F acts on an object of mass m, the acceleration a that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force. ...
... When a net external force F acts on an object of mass m, the acceleration a that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force. ...
2013
... 3. Under what conditions on ma and ms will the equations of motion you derived in questions (1) and (2) above become the same? 4. Determine the components of the absolute angular velocity vector of the satellite ω = ω1 b̂1 + ω2 b̂1 + ω3 b̂1 in terms of quantities given in Figure 1 and your answers f ...
... 3. Under what conditions on ma and ms will the equations of motion you derived in questions (1) and (2) above become the same? 4. Determine the components of the absolute angular velocity vector of the satellite ω = ω1 b̂1 + ω2 b̂1 + ω3 b̂1 in terms of quantities given in Figure 1 and your answers f ...
forces
... If force increases, what should happen to acceleration? If mass decreases, what should happen to force? ...
... If force increases, what should happen to acceleration? If mass decreases, what should happen to force? ...
Static Friction
... •Draw each object separately •Draw all the forces acting on that object •Get x and y components of all the forces to calculate the net force •Apply Newton’s second law to get acceleration •Use the acceleration in any motion analysis and establish a Kinetic Diagram ma ...
... •Draw each object separately •Draw all the forces acting on that object •Get x and y components of all the forces to calculate the net force •Apply Newton’s second law to get acceleration •Use the acceleration in any motion analysis and establish a Kinetic Diagram ma ...
Notes on Terminal Velocity and Simple Harmonic Motion – Physics C
... At t=0, the acceleration is g, since the velocity is zero. This can be confirmed using Newton’s Second Law, mg-kv=ma, and mg-0=ma so therefore a=g at t=0. After a long time, the acceleration of the object is zero. It is important to help the students learn to plug in these limiting values to determi ...
... At t=0, the acceleration is g, since the velocity is zero. This can be confirmed using Newton’s Second Law, mg-kv=ma, and mg-0=ma so therefore a=g at t=0. After a long time, the acceleration of the object is zero. It is important to help the students learn to plug in these limiting values to determi ...