Recent content by e-zero

  1. E

    Masses in contact with each other

    No, and hence net force cannot be the same. I see! :)
  2. E

    Masses in contact with each other

    If 3 masses are in contact with each other I understand that the acceleration is the same for all if a force is applied to the first mass. Is the net force on each object also the same? Assume no friction.
  3. E

    Frictionless incline and required force

    So, in this example, if you we're applying the required force to push it up hill how would you bring the box to rest?
  4. E

    Frictionless incline and required force

    You say that once it's in motion you only have to apply the same force, but wouldn't that just keep the box at rest on the incline?
  5. E

    Frictionless incline and required force

    Oh right. Ok, now am I correct in saying that that opposing force has to be 'slightly greater' in order to push the box up the incline?
  6. E

    Frictionless incline and required force

    I have figure that out, and the only force acting, in this case, is gravity. So that would mean I would need a force mgx in the opposite direction but slightly greater correct? (mgx is the x-component of mg)
  7. E

    Frictionless incline and required force

    If I had to push a box with a certain weight 'm' up an incline, would I just need a force 'slightly greater' than 'mg'? (Assuming I applied the force parallel and up the incline)
  8. E

    Mass gets cancelled out on downhill slope

    Actually after looking at the examples again, I kinda of get it and I realize the ideas and formulas in the examples is what I will be learning more of in the following chapters.
  9. E

    Mass gets cancelled out on downhill slope

    Ok, I might as well post the question: A skier has begun descending a 30deg slope. Assuming the coefficient of kinetic friction is 0.10, calculate the skier's acceleration. The diagram shows that the 30deg angle is at the bottom of the slope where the slope meets the ground. Forumlas for...
  10. E

    Mass gets cancelled out on downhill slope

    I'm looking at a general problem in which a skier is going downhill. I understand 'mathematically' how the mass will cancel out of the final equation when calculating the acceleration, but I don't understand this 'conceptually'. Can anyone describe?
  11. E

    Bucket being lowered vertically by a rope

    How could mg = Ft if it were lowered at a constant speed?? That would indicate that the bucket is at rest.
  12. E

    Bucket being lowered vertically by a rope

    I would say that if its lowering at a constant speed then there is no acceleration. Also, I would say that, if anything, since its moving downwards it would be decelerating. Which I guess is the same as accelerating upwards??
  13. E

    Bucket being lowered vertically by a rope

    Homework Statement A 12.0kg bucket is lowered vertically by a rope in which there is 163N of tension at a given instant. What is the acceleration of the bucket? Is it up or down? Homework Equations Ft - mg = ma The Attempt at a Solution I solved for 'a' and got a = 3.78 m/s^2...
Back
Top