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Static friction and work

  1. Jul 30, 2005 #1
    Could someone please clarify this problem for me...

    Assume we have a box sitting on the back of a truck. The box can be modeled as a particle, and there is friction between the box and the surface of the truck. Also, lets assume the truck is accelerating slow enough that the block does not slide - IE it remains stationary relative to an observer fixed in the frame of the truck. If we draw the FBD, we have two forces in the vertical direction. The weight and the normal which are obviously equal in magnitude and opposite in direction. The only horizontal force we have acting on the block is static friction. In this example we can use the simple definition of work, being the product of force and displacement. However, I don't believe its necessary correct to say that the static friction is doing work on the box, because technically the displacement should be zero, as the static friction causes no displacement of the block relative to the frame in which it is being applied. So is it more correct to say the truck (I guess the engine or the torque on the wheel) does the work, or does the static friction do the work. We had a big debate about this a few months back at one of the TA meetings. I just wondered what your input was....

    One analogy made was something along the lines of how magnetic fields do no work on charged particles, yet they cause the particles to accelerate.

    Thanks to anyone who responds with some sort of plausible argument!
  2. jcsd
  3. Jul 30, 2005 #2
    In total you have FOUR forces acting on the box: two vertical ones and two horizontal ones. The horizontal ones are friction and the force coming from the fact that the box is moving foreward with the truck (= mass_truck * acceleration of the truck).

    Now, in your text you say, ok the box does not move on the truck. This implies that the friction force must cancel out the force that pulls foreward the box (ie the force coming from the truck, so the force exerted by the truck on the box : this is the 4th force you forgot.) so that in total, in the horizontal direction, the acceleration is zero. This gives a constant velocity and since the initial velocity was zero, it stays zero.

    That is correct, friction does not do any work here.

  4. Jul 30, 2005 #3

    Doc Al

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    Staff: Mentor

    The only horizontal force on the accelerating box is static friction. From the inertial frame of the ground, the box is most certainly displacing, so yes the static friction is doing work on the box when viewed from that frame. (Looked at from the non-inertial frame of the accelerating truck, there is no displacement and thus no work. But F does not equal ma in that frame.)

    But the friction is a passive force, not an active source of energy. (It is only transferring energy created elsewhere. Much like a person accelerating a box by pulling on an attached string: the tension in the string is what is literally doing work on the box, but the source of the energy is not the string, but the person.)

    More interesting is the fact that the only force (ignoring air resistance) accelerating the "truck + box" is the friction of the road against the tires. But that force does no work since the displacement is zero. The "ultimate" source of the increasing KE of the truck and box is the chemical energy "stored" in the gasoline + oxygen. That energy is transformed into KE (and thermal energy).
  5. Jul 30, 2005 #4
    Are you sure its correct to say that there are four forces acting on the block. Inertial forces that arise due to the kinematics of a system are often not included in the FBD or are considered physical forces acting on an object. It would be incorrect to say that for a disk whirling in a horizontal circle that is attached to a string to have the EOM... T + m*r*omega^2 = 0
    Rather, the right hand side of the equation is the side containing the kinematics of the problem, and the sum of physical forces acting on the body should be equivalent to the forces produced by the inertial forces (IE centripetal force in this case). In the equation above we would obtain
    T = -m*r*omega^2. This would state that if we defined positve to be radially inward, that the wire would actually have to be in compression as it is producing a radially outward force.
  6. Jul 30, 2005 #5
    That fourth force indeed arises because of the motion of the reference frame of the block with respect to the 'inertial frame'. So, the frame of reference of the block is NOT an inertial frame. this force, will however have an influence on the objects kinematics with respect to the inertial frame. Without wanting to engage in a semantics discussion, i should add that the inertial frame IS the preferred reference frame, keeping that in mind there are indeed TWO horizontal forces acting on the block. But indeed, in the frame of the block, you have only one horizontal force and three forces in total in your FBD, that is correct

  7. Jul 30, 2005 #6
    Just to elaborate my point: suppose we replace the block by YOU on that truck. You know the truck is going to accelerate, so you know that there is a force acting on you. However, you feel pushed backwards (like being in the seat of an accelerating car) because of Newton's third law. This pushing back is an example of these famous fictitious forces. Me, looking at you and the truck from the side of the street (ie the inertial frame) knows that there are two forces acting on you, both equal in magnitude but opposite in signs. However from my point of view, when it comes to dynamics, i only know there is friction acting on you, so one force. I will only see you accelerating with the car. That is what i first meant in my post to you.
    From your point of view, during the acceleration of the truck, you think there is a force pushing you back. Friction keeps you on your position. So in your eyes you would write (you stand still with respect to the truck) 0 = friction - fictitious force. I would write Ma = -frictionforce. Ma is the force associated with the acceleration of the truck and the opposite of this is the fictitious force.

    Now, let us take a little curve with the car. You will be pressed against the door during the curve. From the general reference frame outside the car it is clear that there is a pair of equal and opposite forces acting between the passenger and the car. From the specific reference frame inside the car, centrifugal force exists as a fictitious force which appears to try to eject you from the car as the door pushes on you to counteract your inertial tendency to keep moving in a straight line at a constant speed. We feel forces acting on us in the opposite direction from the actual forces that are causing us to accelerate.

    Last edited: Jul 30, 2005
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