Houston…we have a problem 1) You learned that if you push on a door far away from it’s hinges...

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Transcript of Houston…we have a problem 1) You learned that if you push on a door far away from it’s hinges...

Page 1: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.
Page 2: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

Houston…we have a problem

http://www.youtube.com/watch?v=eco_xvkEQlg

1) You learned that if you push on a door far away from it’s hinges (the axis of rotation) it will be easier to open and close

2) You learned if you hang a mass farther away from your hand (the axis of rotation) it was harder to move up and down

Don’t they contradict one another?

Page 3: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

So what exactly did we learn?The direction that you apply the force is

rather important

If you apply your force in the same direction as the object moves… less effort is needed with a longer lever

If you apply your force opposite the direction that the object moves…more effort is needed with a longer lever

Page 4: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

What is torque?

Torque is a tendency of a force to rotate an object about an axis

A force applied to an extended object can produce a torque.

This torque in turn causes the object to rotate.

Page 5: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

The object rotates about an axis.

The lever arm is the distance (in meters) from the axis of rotation to the point where the force acts.

The applied force is the force applied to the lever arm.

Page 6: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

Formula For Calculating Torque

Torque = (force perpendicular to axis)(length of lever arm)

τ = Fl d

τ = the Greek Letter Tau

Fl = Force perpendicular to lever arm

d = length of lever arm.

Page 7: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

Calculating torque

The torque (τ) the Greek Letter Tau, created by a force is equal to the lever arm (r) times the magnitude of the force (F).

Page 8: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

You try…A force of 50 newtons is applied to a wrench

that is 30 centimeters long.Calculate the torque if the force is applied

perpendicular to the wrench so the lever arm is 30 cm.

Page 9: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

You try…

τ = r x F = 0.3 m x 50 N = 15 N-m of

Torque

Now lengthen the arm of the wrench to 0.7 m

Page 10: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

You try…If the doorknob was

placed in the middle of a door, rather than at the edge..how much more force would be needed to produce the same torque for opening the door?

Twice as much

Page 11: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.
Page 12: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.
Page 13: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.
Page 14: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

Balanced (Net) TorqueThe net torque is the

sum of all the torques produced by all the forces

Net-torque is zero or.. the torques are balanced

Counterclockwise torque = clockwise torque

Page 15: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

Balanced (Net) Torque

Equation

Fleft x rleft = Fright x rright

Page 16: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

Calculate using equilibriumA boy and his cat sit

on a seesaw.The cat has a mass of

4 kg and sits 2 m from the center of rotation.

If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?

Page 17: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

A boy and his cat sit on a seesaw. The cat has a mass of 4 kg and sits 2 m from the center of rotation. If the boy has a mass of 50 kg, where should he sit so that the see-saw will balance?

Page 18: Houston…we have a problem  1) You learned that if you push on a door far away from it’s hinges (the axis of.

Less force is required farther out from the center of mass than closer in. So, by tackling a runner low -- far from the center of mass -- it takes less force to tackle him than if he were tackled high. Furthermore, if a runner is hit exactly at his center of mass, he will not rotate, but instead will be driven in the direction of the tackle. Similarly, coaches often advise linemen to stay low. This brings their center of mass closer to the ground, so an opposing player, no matter how low he goes, can only contact them near their center of mass. This makes it difficult for an opposing player to move them, as they will not rotate upon contact. This technique is critical for a defensive lineman in defending his own goal in the "red" zone, the last 10 yards before the goal line.