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Forces, velocities, and angles

  1. Oct 29, 2006 #1
    I have a take home quiz for my physics class. Out of the 10 questions I have answered 7 of them, whether they are right I don't know. First and foremost I need help on the ones I haven't solved yet. I am not necessarily looking for someone to give me the answers, but rather help me setup the problem and use formulas to arrive at the answer myself.

    The hardest question is #3.

    A 200lb man stands tiptoed on one foot so that all his weight is borne by the ground beneath the ball of the foot. If the foot and ankle are considered as an isolated body, the three forces that are in equilibrium are the reaction W of the ground, the pull T of the achillies tendon and the compression C of the tibia. The force C is downward at an angle 15* from vertical. The force T is upward at 21* from the vertical. W is an upward force of 200lbs. Calculate the values of C and T. (HInt: one way to do this is to have two simultaneous equations, one balancing the horizontal components of force and the other balancing the vertical components. Another way is to use the law of sines)

    #2.

    A rocket exhausts fule with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with the fuel comprising 80% of the total mass. When all the fuel has been exhausted, what is the rockets speed?

    #5.

    A car rounds a 75m radius curve at a constant speed of 18m/s. A ball is suspended by a string from the ceiling of the car and moves with the car. What is the angle between the string and the vertical (y-axis)?
     
  2. jcsd
  3. Oct 29, 2006 #2
    the first problem is not nearly as difficult as it looks. The only problem is that you are saying things like "upward from verticle, which actually does not make to much sense. Sorry. I'll help you if you explain the directions.

    for the second problem, you know that momentum is conserved. If the rocket fuels final momentum is (1500 m/s)(4 <arbitrary mass units noting that the fuel has a 4:1 ratio to the ship>) and the ships momentum is
    -(Velocity of ship)(1) then you know that the velocity of the ship equals four times the speed of the fuel.
     
  4. Oct 29, 2006 #3
    Believe it or not the last question is certainly the hardest.

    First, note that the car is accelerating inward. I'm sure you know how to calculate this acceleration, so I'll leave that to you and just call it "a"

    Now forget the circle stuff, and just pretend that the car is accelerating forward. Because the ball has inertia, it resists the acceleration, and therefore makes an angle. Now imagine that acceleration is perfectly constant and the ball has made a beautifully still angle. you know that the ONLY upward force acting on the ball is the y-component of the tension. This component is equal to Tcos@ which must also equal the balls weight mg. So now you know that T=mg/(cos@). Read that again if i lost you. Now pretend that the car is not accelerating, but instead, a "fictitious force" is acting directly backwards on the ball. this force is keeping the angle. you know that there are two forces now acting on the ball in the X direction. Forward is the x-component of tension Tsin@=(mg(sin@))/(cos@)=mgtan@. the other force is the fictitious force balancing out the tension's x-component. This force is (M)(a) <the same "a" as before. so mgtan@=ma so a=gtan@ so @=tan-1(a/g)
     
  5. Oct 29, 2006 #4
    Wow, thanks bro, I'm gonna work those out. As far as the first question, when I say 15* or 21* from vertical, I guess it be easier to describe as 105* and 111* from a horizontal plane. I copied the question exactly, but it had a picture with it, so for me the visualization went along with the picture, I guess without it it's a little harder to visualize. Did that help?
     
  6. Oct 29, 2006 #5
    Okay this is not too hard. I would definitely avoid the law of sines because that just makes it complicated.
    what you can do (as it says) is take two equations--one for the forces in the X direction and one for the Y direction.
    so Fnet(Y)=0 because it is not moving and also Fnet(X)=0
    first let's do Y. Force W must equal the vector sum of the other two forces so that Fnet(Y)=0
    See if you follow this: 200lbs=C(cos15)-T(cos21)
    in that case, the cosine of C and T make the Y components of them and therefore that expresion makes since.
    next let's do X
    W does not matter since it has no X-component
    I am assuming even though you did not say that the 15 degrees from verticle and the 21 are pointing in the opposite horizontal directions.
    C(sin15)=T(sin15)
    make since? Now you just have a system of equations to be solved with substitution.
     
  7. Oct 30, 2006 #6
    Not quite sure what you mean in your last resposne, but both angles are relative to the leg. Both angles fall in the same quadrant(II), how ever the force for the tibia is down and the force of the tendon is up. Thanks again for all the help. I wish I would of founf this place sooner.
     
  8. Oct 30, 2006 #7
    (wow I just got home from school,logged on, and you reponded about one minute ago!)
    Okay, I am slightly confused. I instead of telling me the quadrants, tell me how they point. As in Up-left and Down-Right.
    If it is either that or Up-right and Down-left then my explanation works. Would you like me to explain what I did in some detail?
     
  9. Oct 30, 2006 #8
    The Y forces are as follows. W is just up--nice and easy.
    C points down at an angle, but the Y-component points just down.
    That is C(cos15) is down (that might sound weird for a y-comp. to be cosine but you could also say sin105.
    So you would just say that W=C(cos15) but the force T pulls up. It pulls up with T cos 21 okay?
    So now we have W=C(cos15)-T(cos21)
    (becuase it does not have to pull as much do to T
    Call that Equation one.
    for x components, W does not matter, so you just set the x-comps equal to each other
    T(sin21)=C(sin15)
    Did that clear thing up?
     
  10. Oct 30, 2006 #9
    Yeah it did, thanks. It is weird explaining the diagram without being able to show it to you, but I guess you had it right.

    Okay, so here I go.


    I came up with the values c=-2670.92N and t=-2075.96N For some reason these figures don't seem right, but I'm not sure. I solved for C and T using substitution. I converted the 200lbs for W into kg and then Newton's =892N so

    -2670.92N(cos15)-(-)2075.96N(cos21)=892N
    Just seems a little weird. How far off am I?
     
  11. Oct 30, 2006 #10
    For the fuel question, the ships velocity equals 6000m/s, but I am still having trouble gripping the concept of it. If it's 4 times the fuels velocity. The book says Mvi=Pi
    Pf=(0.80M)v(fuel) +(0.20M)v(ship)
    Vship= Vrel +Vfuel
    Vfuel= Vship - Vrel
    Mvi= 0.80M(Vship - Vrel) + 0.20MVship

    I totally lost.
     
  12. Oct 30, 2006 #11
    The ball.
    Okay, I get the forces acting on the ball. Given that ma=mgtan@
    a/g=tan@
    tan@= 4.32/9.8= 0.441
    tan-1(0.441)= 0.415* angle

    Doesn't seem like much of an angle, but I guess the radius would have to be smaller of the speed quicker to force the ball to move more right?
     
  13. Oct 30, 2006 #12
    Calculator was in radians.
    C=3057.94N T=2209.01N

    Still seems awefully out there.
     
  14. Oct 31, 2006 #13
    calculator was in radians for this one too. I got 23.78*.

    I appreciate all the help. I have to turn it in now, but I know I have done better than I would of without the help. Thanks a lot.
     
  15. Oct 31, 2006 #14
    Well I turned in the exam. I got all of them right except the rocket fuel question. We did not go over it, so I don't know what the answer was, but I am going to check, a class mate found a problem in the book exactly like it, so I'm a plug a chug and see what comes up. He said he got like 24xxm/s so we'll see. Thanks again for all the help, I really appreciate it.
     
  16. Oct 31, 2006 #15
    This answer was correct, however I left of then - for C, I hope I remebered it on the test, otherwise I might lose some points. :eek:
     
  17. Nov 1, 2006 #16
    I found the formula for the rocket and fuel.

    Vf-Vi= Vrel(ln(Fi/Ff) Where F is the fuel initial/final

    Vf-0=1500(ln(1/.2)

    Vf=2414m/s

    I don't know if I followed your method correctly or not, but this is not the answer that I got.
     
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