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Join date : 2010-01-05
Location : British



A block of wood is given a push and slides along a horizontal, flat table surface, coming to a stop. Of course it's obvious that friction is the reason that its kinetic energy drops to zero when it stops, work having been done as a result of the force due to friction at the interface between the table and the bottom of the block.
(1) But what about the block's momentum? What happened to that?

(2) And what about the block's angular momentum? It doesn't have any, you say. Well, about its own center of mass it surely isn't rotating during this experiment. But why not? The only forces on the block are the forces due to friction at its bottom, the gravitational force, and the upward reaction force the table exerts on the block. The net (total) of the gravitational force and the reaction force must be zero, since the block isn't accelerating up or down. The force due to friction acts in the plane of the table and therefore has a lever arm about the center of mass, and therefore must exert a torque on the block about the center of mass. So why doesn't this set the block into rotation? There must be a counter-torque of equal size and opposite direction to the friction torque; but what is it?


(1) Momentum of a system is conserved. The horizontal momentum lost by the block is given to the table through the continuous impulse of the force due to friction at the table-block interface.

(2) This one is more interesting. Take an axis through the block's center of mass, a distance H above the table, where H is the height of the block's center of mass (C) above the table. The force due to friction f supplies a clockwise torque fH about this axis. The gravitational force is balanced about the center of mass, and gives no net torque. We conclude that the normal forces along the bottom surface of the block cannot be balanced about the center of mass, but must be greater at the front of the block, providing the necessary counterclockwise torque to balance the torque due to friction, and resulting in no appreciable rotation of the block about its center of mass.
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Join date : 2010-01-05

PostSubject: Re: PHYSICS PROBLEM CHALLENGES UNDERSTANDING   Tue Jan 05, 2010 11:07 pm

I will add this information too.
Braking automobile.
Automobile brakes are often designed differently for front wheels than for the rear wheels. The front wheels are more rugged, and sometimes are disk rather than drum brakes. Why is this?


When the automobile brakes, you notice that the front dips down. The front springs compress more than the rear ones, and the front tires do also. This indicates there's more vertical force on the front axle than on the rear.

The reason this happens is that the force of friction of the roadway on the tires produces a clockwise torque about the car's center of mass (see picture below). The vertical normal forces on the tires must adjust in size to produce an equal and opposite counterclockwise torque.

In the light of the previous problem of the sliding cube, the reason is clear. The center of mass of the vehicle is between the front and rear axles, and above them. Therefore the force due to friction between roadway and tires provides a torque that initiates a rotation, causing the downward dip of the front of the vehicle. This compresses the front springs more, causing the normal force to be greater at the front than at the rear. This torque is opposite to the torque due to friction, The forces on the tires due to friction f1 and f2 at the roadway are linked to the normal forces N1 and N2 through the relation f = µN where µ is the coefficient of friction. So, since N2 is greater on the front tires than is N2 on the rear tires, the friction at the roadway is also greater on the front tires. The brake drums or disks in the front also have an increased force due to friction, hence they must be more rugged.
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