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Help on a few more?

  1. Nov 29, 2004 #1
    Help on a few more??

    Hi guys,

    I still need some help with physics. If you guys could help me get a good grade, it would be soooooooooo appreciated.

    Okay, this one I've been working at for a while:

    A 17.0 kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 220 N and breaks as the elevator accelerates. What was the elevator's min acceleration (magnitude & direction?)

    This one too:

    What is the smallest raduis of an unbanked curve around which a bicyclist can travel if her speed is 18 mph and the coefficient of static friction between the tires and the road is .32?

    One last one:

    A get palne starts from rest of the runway and accelerates for take off at 2.3 m/s^2. It has two jet engines, each of which exerts a thrust of 1.4 x 10^5 N (approx 16 tons) What's the weight of the plane??

    If you guys could help, I would be ETERNALLY grateful!! Thanks so much for helping me!!!!
     
  2. jcsd
  3. Nov 29, 2004 #2
    For the first and last ones: draw a freebody diagram, label all the forces, then use F=ma to find what they want to know.

    For the second one: draw a freebody diagram, label all the forces, then use the centripital force equation F=(mv^2)/r to find the radius.

    Hint on the second one: the friction force will be directed towards the center of the circle, while the centripital force will be directed outwards.
     
  4. Nov 29, 2004 #3
    Thanks, but wait, I'm still a little confused on the bike problem. We need to know the mass of the bike and the rider, and that's something we don't know yet. How do you find the mass? wait...tell me if I'm doing this right.....

    g=gravity? So to find the mass would you use the equation g=G*me/re^2?? If not...still, how is the mass found??
     
  5. Nov 30, 2004 #4
    This is wrong. The centripetal force and the friction force are both directed towards the centre of the circle. In fact, the friction force provides the centripetal force, confining the bicyclist to a circle.
     
  6. Nov 30, 2004 #5
    Hi,

    All these problems may be solved by using Newton's second law. You begin by writting out [itex]\Sigma F & = ma[/itex] and the rest is just mathematics.

    For the monkey
    [itex]
    \begin{align*}
    \Sigma F & = ma \\
    T - mg & = ma\\
    T & = ma + mg
    \intertext{string breaks when}
    T & \geq 220N \\
    \intertext{so}
    ma + mg & \geq 220N \\
    ma & \geq 220N - (17.0kg)(9.8ms^{-2}) \\
    a & \geq 3.14 ms^{-2}
    \end{align*}
    [/itex]
    so the minimum acceleration is [itex]3.14 m s^{-2}[/itex] upwards.

    The bicyclist is moving in a circle at constant speed so we know that the sum of the forces acting on her must be towards the centre of the circle and of magnitude [itex]mv^2/r[/itex]. Yes [itex]\Sigma F_r & = ma_r[/itex] is still true, but it's more interesting to write [itex]\Sigma F_r & = mv^2/r[/itex].

    For the bicyclist
    [itex]
    \begin{align*}
    \Sigma F_y = ma_y \\
    \intertext{there is no vertical motion so}
    n - mg & = 0 \\
    n & = mg \\
    \Sigma F_r & = mv^2/r \\
    \intertext{the vertical forces are balanced so the only force providing the centripetal force is the static friction with the road.}
    f & = mv^2/r \\
    \intertext{If you look in your textbook you'll find that}
    f & \leq \mu_\mathrm{s}n \\
    \intertext{so}
    mv^2/r & \leq \mu_\mathrm{s}n \\
    mv^2/r & \leq \mu_\mathrm{s}mg \\
    \end{align*}
    [/itex]
    [itex]r & \leq \frac{v^2}{g\mu_\mathrm{s}} [/itex]
    Now just plug in the numbers, remembering to convert 18 mi/h into units of m/s.

    For the aeroplane
    [itex]
    \begin{align*}
    2\times 1.4 \times 10^5 N & = m(2.3ms^{-2})\\
    W & = mg
    \end{align*}
    \end{align*}
    [/itex]
     
    Last edited: Nov 30, 2004
  7. Nov 30, 2004 #6
    Nope. I think you meant centripetal acceleration. The centripetal acceleration is directed towards the center of the circle but the force produced by the motion (centripetal FORCE) is directed outwards.

    If you drive your car around a curve, don't you feel a force pushing you to the outside? That's the force I'm talking about.

    One more thing, if both forces are towards the center of the circle, then why does the bicyclist not just move towards the center?
     
  8. Nov 30, 2004 #7

    arildno

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    jdstokes is correct:
    We use the "centripetal" concept to mean towards the center, "centrifugal" away from the center.
    I haven't checked calculations in this post, though.
     
  9. Nov 30, 2004 #8

    ZapperZ

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    This is not correct. Since F=ma, where F and a are both vectors, the direction of a and F must be the same. So already by saying that the centripetal acceleration in inwards, but the centripetal force is outwards is not correct. The centripetal force must always be inwards, the same direction as the acceleration.

    Secondly, when you drive around a curve IN the car, you are experiencing the fictitious "centrifugal force". This is because by being IN the car, you are no longer in an inertial reference frame. In fact, when the car in going in a circle, you are in an accelerated reference frame. In this frame, there will appear to be "forces" that actually are not present (or not in the "right" direction) when viewed in an inertial frame. The outward force you are experiencing is due to a combination of two factors: (i) the tendency of your body to want to move in a straight line and not change direction; and (ii) the pull of the side of car directing you inwards so that you follow the trajector of the car that is going in a curve.

    If you look from the OUTSIDE of the vehicle in a proper inertial frame, you will see the car (and its content) moving in a circular path, and the only force acting on it to make it move that way is the centripetal force provided by the friction between the wheels and the road surface. For the system, this is the only "true" external force that is causing the vehicle to move in the circular path. You do not have the outward centrifugal "force" anymore.

    Zz.
     
  10. Nov 30, 2004 #9

    arildno

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    It is unfortunate, IMO, that the concepts "centrifugal", "centripetal" are used in classifying of forces/pseudo-forces.
    These concepts are intimately related to ACCELERATIONS, and a "centripetal force" doesn't mean anything else that "some force providing centripetal acceleration".

    That is, the concepts provide no further information of the type/nature of the forces involved (what mechanism which generates them, for example as being a type of friction), and hence, IMO, they are not a particularly interesting way of classifying forces.
     
  11. Nov 30, 2004 #10

    ZapperZ

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    But you could say that with ALL the forces we deal with. They are, after all, very tied to the fact that you have a "test" object, such as a mass, or a charged particle, and then detecting a dp/dt on that test object. Only then are you able to defined a "force". So saying that a centripetal force is ANY force that produces the centripetal acceleration for an object to move in a circular path is no different than what we can say about any other forces.

    I don't think "centripetal force" is a classification of forces. In other words, we are not saying the different types of forces are gravitational, electrostatic, centripetal, etc... This would be incorrect, and I don't think any textbook tries to do that. It is incumbent upon the instructor to clearly state that a "centripetal force" is some generic term given to any force that can provide a circular motion. I don't think this would hurt anything. In fact, the whole concept of "force" itself is generic, and the student must quickly grasp that understanding or else he/she would think one needs a new set of mathematical tools when a switch is made from "mechanics" forces to electrostatic/magnetostatic forces.

    Zz.
     
  12. Nov 30, 2004 #11

    arildno

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    "In other words, we are not saying the different types of forces are gravitational, electrostatic, centripetal, etc... This would be incorrect, and I don't think any textbook tries to do that."
    Precisely my point.

    However, I've encountered several students who search for the "centripetal force" in addition to the gravitational force, (or normal force in some other situations) acting on an object.

    I'm rather tired of this misconception, possibly students in other parts of the world never fall prey to that fallacy.
     
  13. Nov 30, 2004 #12

    ZapperZ

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    I have encountered that too. That's why I think textbooks, and especially instructors, should be explicitly clear that the "centripetal force" is generic and can be of any source, be it "tension in the string", electrostatics, magnetostatic, friction, etc... But to be fair, none of them ever stated "centripetal force" as a "type" of force. So at some point, we need to whack the back of the heads of these students to wake them up!

    :)

    Zz [who has never been accused of student abuse EVER!] :)
     
  14. Nov 30, 2004 #13

    arildno

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    My personal preference is for whipping their butts, but head-whacking is kind of cool, too..:wink:
     
  15. Nov 30, 2004 #14
    I apologize if I were wrong but it's really not my fault. I took mechanics last year and my physics teacher always taught us that the centripetal force is outwards and the friction is always in the direction that opposes the force.

    I will read online about that subject right now and make sure I understand all of it because I feel really bad not knowing how to draw a freebody digram for a car going around a circle (well I do using what my physics teacher told me but it's apparently wrong).
     
  16. Nov 30, 2004 #15

    ZapperZ

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    If what you said is true, then it appears that your teacher also needs a head-smacking and butt-whipping. Both arildno and I would gladly volunteer to do that.

    If you look at what you just said, you can immediately notice that there's something not kosher there. If there really is a "centripetal force" acting outwards, and a friction force acting inwards ".. that opposes the force", then this means that the NET force on the object is ZERO! Now you should realize from Newton's First Law (which is NOT a metaphysics) that under such circumstances, the object will only move in a straight line, or not move at all! This would be weird since we ARE talking about an object in a circular motion, and thus, it HAS an acceleration and therefore MUST have a net force. So if you just think about it a little bit, what you just said and have been taught makes no sense.

    Zz.
     
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