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Work problem

  1. Oct 15, 2006 #1
    "Work" problem

    I'm sure this is a straight forward problem, but I think that I may be taking the wrong approach. Any guidence would be appreciated.


    In many neighbourhoods, you might see parents pulling youngsters in a four-wheeled wagon. The child and the wagon have a combined mass of 50 kg and the adult does 2.2*10^3 J of work pulling the two for 60 m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.

    a) Draw an FBD for the wagon (Not a problem)
    b) Determine the magnitude of the force applied by the parent.
    c) Determine the angle at which the parent is applying this force.

    For b) Since speed is constant F(net) = 0.
    Therefore, F(a) - F(f) = 0, F(a) = F(f)
    F(f) = u(k)F(n)
    = 0.26((490 N)
    = 127.4 N

    So...the magnitude of the force applied by the parent is 127.4 N?

    For c) Since W = Fd
    2.2*10^3 = 127.4 N cos theta (60 m)
    = 7644 J cos theta
    cos theta = 2200 J/7644 J
    = 0.2878
    theta = 73.3 degrees.

    So...the angle at which the parent is applying this force is approximately 73.3 degrees?
  2. jcsd
  3. Oct 15, 2006 #2


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    If the parent is pulling at an angle, the normal force is not the weight of the wagon and child being pulled. Look at your FBD and think about the normal force.
  4. Oct 15, 2006 #3
    I see what you're saying OlderDan, thank you for pointing out that error. I'm thinking F(n) should be (mg - F(a) sin theta). Do I not need the value of theta in order to do anything with this? Every way I think about this question brings me deep into the seedy underworld of theta. I can't seem to determine it's value with the information given. What am I missing here?
  5. Oct 15, 2006 #4
    A little clue around here.. If wagon is moving at constant speed, then force exerted by parent that causes ur wd will be ur kinetic friction. With the value of kinetic friction, u will be able to find the value of the normal force.
  6. Oct 15, 2006 #5


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    Read gunblaze post #4. Remember that work is the force times the distance in the direction of the force, or the component of force in the direction of motion times the distance. That gives you information about the force component in the direction of motion.
  7. Oct 21, 2006 #6
    One more try here...

    Since, F(a) = -F(f)

    Can I use F(f) = (W = Fd)
    2200 J = F(60 m)
    F(f) = 36.7 N (or -36.7 N)

    Since, F(f) = u(k)F(n)
    36.7 N = (0.26)F(n)
    F(n) = 141.15 N

    Since acceleration = 0, then F(net) = 0. Would F(a) = F(n)???

    Am I still lost here? This question seems as though it should be relatively simple enough.
    Last edited: Oct 21, 2006
  8. Oct 21, 2006 #7


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    The normal force looks correct, but it is substantially less than the weight of the child and the wagon. What force accounts for this difference?
  9. Oct 21, 2006 #8
    The vertical component of the force applied by the parent. Which will be 348.85 N (mg - F(n)). The horizontal component of force applied by the parent will equal F(f) since there is no acceleration, which is 36.7 N. With these values I can determine the net force applied by the parent to be 350.78 N at an angle of 84 degrees. Anybody?
    Last edited: Oct 21, 2006
  10. Oct 21, 2006 #9
    Don't skip steps.

    Write down the equations you're using, and what you know:

    Weight = Mass*Gravity Constant
    Normal Force = -[Wieght + ParentForce*sin(theta)]
    Friction Force = Normal Force * Mu
    Summation of Forces = Mass * Acceleration
    Work = Force*Displacement*cos(theta)

    (Keep in mind, the type of work you're talking about only exists on the X-axis where the frictional force is. Sine is for Y-axis scale changes, not X-axis.)

    Work = 2.2x10^3 Joules
    Mu = 0.26
    Distance = 60 meters
    Mass of system = 50 kilograms

    Okay, so you are looking for the force vector of the parent. Now, I do not know how your teacher wants you to find the force before the direction, but I found the aswer in the opposite way. (Only took half a sheet of paper, too) Let us start with the basic plug & chug equations:

    Weight = M*g = 50kg*9.806m/s/s = 490.3 N

    Okay, after this it gets rather complex if you are not used to having two variables (In this case, F*cos(theta)) together equaling something, and substituting it in for other variables. We need to solve for F*cos(theta). We can do this with the simple work formula and algebra.

    Work = Force*Displacement*cos(theta)
    Work/Displacement = Force*cos(theta)
    36.66 (repeating) N = Force*cos(theta)

    We know both the values for Work and displacement, therefore we can substitute that value for anywhere we see "Force*cos(theta)". Like in the next system.

    In the free body diagram, we have an interesting dynamic: the parent's force is actually reducing the amount of normal force because she is not only pushing the wagon forward, but upward on a vector. As you know, normal force is the negative summation of y-axis forces (Depending, of course, on where the forces are being applied. In this case, it is the ground, so we're on the Y-axis). Therefore, our equation for normal force must look like this:

    Normal = -[Force*sin(Theta) + Mg]

    (Mg pushing down, Force*sin(theta) pulling up)

    The reason we needed the normal force was, obviously, to find the frictional force. The frictional force is the normal force multiplied by Mu, in this case it is 0.26. Now, if we summate the two forces, they must equal zero since there is no net acceleration.

    Pulling Force + Friction force = 0
    Force*cos(theta) + -(.26)[Force*sin(Theta) + Mg] = 0

    Let's plug in our values that we've come up with so far:

    36.66 N - .26*Force*sin(theta) + 127.478 N = 0

    Let's do some algebra. Subtracting both sides the two like terms to the other side and dividing both sides -.26, we get:

    Force*sin(theta) = 631.32564 N

    Now, you must remember your trigonometry formulas and remember a little bit of arithmetic:

    Identity property: Dividing two numbers by themself equals 1.
    Tan(theta) = Sin(theta)/Cos(theta)

    Knowing this, we can divide our two numbers together to give us a strait number to use algebra with to find our direction:

    Force*sin(theta)/(Force*cos(theta) = Tan(Theta) (Forces would have cancelled out)

    Tan(theta) = 631.325664 N/36.6666666 N = 17.21797
    Theta = Archtan(17.21797) = 1.512782694 Rad

    That's your direction. Now for your magnitude. We need only go back to our previous statement of:

    Force*cos(theta) = 36.666666 N

    Now just plug and chug.

    Force = 36.6666666 N/cos(1.512782694 Rad) = 632.38952 N

    I would check it, but I've already spent half an hour helping you. Good luck in physics. Are you taking it in high school or college?
  11. Oct 21, 2006 #10
    I second in not skipping steps, if you show everything not only can you see where you went wrong, but on tests you may get partial creidt. Just a thought.

  12. Oct 21, 2006 #11


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    I think you got it. The angle seems afwfully high, but this is what the numbers give you.
  13. Oct 21, 2006 #12


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    I think it needs to be checked. I have not had time to check it carefully, but I think you have a sign error here
  14. Oct 22, 2006 #13
    Yeah, you're right there is an error. It's odd though, because that shouldn't happen. Alas, it has been a year since I've studied FBD and dynamics...

    This is a strange problem. My assumption shouldn't have distorted the answer, but it did.

    The normal force always pushes in the opposite direction of the net forces pushing down. In this case, the weight is being supported by not only the normal force, but by the parent's vertical force and the weight.

    Well, let's take a mathematical looksee at this again:

    Normal Force + Force*sin(theta) + M*g = 0 Because the Y-axis forces have to be at rest.


    Normal Force = -Force*sin(theta) + -M*g => -[Force_sin(theta) + M*g] (factoring out a -1)

    That should work, should it not?

    EDIT: Okay, let's look at this again. This time we'll look strictly at the normal force and it's relation to the frictional force. We'll divide the force formula by .26 to get our values for the normal force:

    141.92564 N + -Normal Force = 0
    Normal Force = 141.92564 N (Subtracting the constant, and dividing by -1)
    [Force*sin(theta) + -490.3 N] = 141.92564 N (Substituting our other value)
    Force*sin(theta) = 632.22564 N

    Yup, there's something wrong about this equation, but for the life of me everything looks fine.
    Last edited: Oct 22, 2006
  15. Oct 22, 2006 #14


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    You always have to be careful about the signs and be consistent in their application. If you assume the applied force is positive, with two positive components, and you take N as positive, and you take g as positive, and you take upward as positive, then

    N + Fsin(θ) - mg = 0
    N = mg - Fsin(θ)

    The parent is pulling upward on the sled, opposing gravity and reducing the normal force required to support its weight.

    In your equation, something has to be negative. That would be OK if you consistently used it as negative in your other equations and calculations.

    It should have tipped you off that you had an error in your calculation when you found that the vertical component of the applied force was greater than the weight of the object that was being towed.
  16. Oct 22, 2006 #15
    This is a grade 12 advanced correspondence course, it's ultra condensed, poorly presented and getting assistance from a teacher is an obstical course of sorts. This is why I've nominated the fine folks here at physicsforums my honorary teachers. Thanks for the help on this one, I'm actually quite glad to see that others had difficulty with it (It's not just me). So...I'm going with the solution I've arrived with in post #8.
  17. Oct 22, 2006 #16


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    Good choice.
  18. Oct 26, 2006 #17
    Yes, which was my tip off that I did something wrong... However, what was wrong with my calculations? Where did I change my sign convention?
  19. Oct 26, 2006 #18


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    There is no way for me to tell that without seeing every substitution you made of a numerical value for a variable. I am not inclined to do that even if I could see it. I suggest you do the problem again adopting a convention that all magnitudes of all variable forces are positive, and use - signs as needed when forces are opposing.
  20. Mar 23, 2007 #19
    Hi, I am doing the same correspondence course and am at this question. I don't understand why Fn would not be equal to -Fg. The wagon is being pulled horizontally so why would the Fn be different?
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