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I need help with some fisix questions RE: Forces and Torque Equilibrium

  1. Apr 17, 2006 #1
    Hey there, we are working on a unit in fisix on forces and torque equilibrium and right now I am stuck on some questions and I have no idea how to solve them, hence the reason I am here.

    Question 1:
    http://www.elarune.net/admins/josh/q21.jpg

    Again, I draw a blank and have no idea on how to solve this question.

    Question 2:
    http://www.elarune.net/admins/josh/q24.jpg

    For this question I know I would have to use vector additions. I think I would have to first just use T1, T2 and the Fg of that 200N ball. But then looking at the angles and such, I start to get confused and it seems to not work in my mind and then I could not find T3. So I also have no idea what to do.

    Final Question.
    We did a lab the other day in fisix where we had a 2x4 on a hinge on the wall an we had a rope with a scale on it to mesure the tension. Under the 2x4 we had two masses hanging down different distances from the pivot point. From that we had to find the mass of the 2x4. I was able to do it. Now in the lab write up, the teacher wants us to create a vector diagram with the vorces of Fg, Fn, T, F1 and F2. He says that Fg, F1, and F2 would create a line straight down, Fn would be perpindicular to that and then the T would finish up the triangle to make a right angle triangle. Well, I am doing this and I get a triangle, but it is not a right angled one and it does not look like his. First I did the triangle with forces and then I did it again with Torque but I still was not able to get the triangle to look similar.

    Thank you all for all your help.
     
    Last edited: Apr 17, 2006
  2. jcsd
  3. Apr 17, 2006 #2
    I bet if you spent the time it took posting your entire homework set trying to solve it, you could have solved it.
     
  4. Apr 17, 2006 #3
    Nope. I spent three days looking at those questions and about 20 minutes typing all of that up. These are only a small portion of all the questions that we have been assigned.
     
  5. Apr 17, 2006 #4
    No, I actually want to learn how to do these questions. I'd rather have someone give me suggestions on how to solve them, than actually telling me how to do it.
     
  6. Apr 17, 2006 #5
    for question 6 sum the torques and let me know what you get. this is the only one im sure of how to do
     
  7. Apr 17, 2006 #6
    For number one, consider the tension (T) in the string. The cosine of the angle times the tension (T) must be equal to the force due to gravity (this is a statics prob). The sine of the angle times the tension must be equal to the acceleration of the car.
     
  8. Apr 17, 2006 #7
    For number two, the pullys (pullies?) redirect the tension forces (is that a foot?). The tension in the string is the same everywhere throughout the string. So, 2 times the cosine of the angle theta times the tension is equal to the force exerted by the foot.
     
  9. Apr 17, 2006 #8
    Question 6:

    Ok, so the torques would be from his Fg, Fg of the ladder and the Ff, would I go:

    Tccw = Tcw
    60kg(9.8m/s/s/)(1m)sin30 + 5kg(9.8m/s/s)(2m)sin30 = Ff(4m)sin60
    294Nm + 49 Nm = Ffsin60
    343 Nm = Ff(4m)sin60
    Ff = 99 N

    Question 1:
    Tcos45 = Fg
    T = Fg / cos45
    T = (2kg)(9.8 m/s/s) / cos45
    T = 13.86 N

    Tsin45 = acceleration
    13.86 sin45 = acceleration
    9.8 m/s/s = accerlation.

    If this is correct, then the way I did it originally must be correct.

    For question 2

    2cos(Theta)T = Ffoot

    so I assume that the tension is equal to Fg? Meaning that the equation would be

    2cos(60)(21kg)(9.8m/s/s) = Ffoot
    206 N = Ffoot?

    Thank you so much so far.
     
    Last edited: Apr 17, 2006
  10. Apr 17, 2006 #9
    For question 5 part (b), i think they want a force equal in magnitude but opposite in direction. Meaning, the x and y components of the force you come up with are equal to and opposite in direction of the of the sum of the individual x and y components of the forces in the diagram.
     
    Last edited: Apr 17, 2006
  11. Apr 17, 2006 #10
    For question 7, if you made point B the pivot i think it would work better; you trying to find the torque (they want the force though-which is the torque divided by distance) at A.
     
    Last edited: Apr 17, 2006
  12. Apr 17, 2006 #11
  13. Apr 17, 2006 #12

    arildno

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    Ok, so the torques would be from his Fg, Fg of the ladder and the Ff, would I go:

    Tccw = Tcw
    60kg(9.8m/s/s/)(1m)sin30 + 5kg(9.8m/s/s)(2m)sin30 = Ff(4m)sin60
    294Nm + 49 Nm = Ffsin60
    343 Nm = Ff(4m)sin60
    Ff = 99 N


    You cannot do it like this, because there will also be a NORMAL FORCE acting upon the ladder from the ground, creating its own torque about the ladder top.

    The simplest way to do this problem, is to utilize the condition that the wall is SMOOTH.
    That means that the wall exerts only a normal force, i.e, in the horizontal direction.
    The easiest way to proceed is to eliminate the unknown ground force by computing the torque with respect to the ladder's contact point on the ground.

    Since the system is horizontally at rest, the frictional force from the ground must balance the wall's normal force.
     
    Last edited: Apr 17, 2006
  14. Apr 17, 2006 #13
    But then wouldn't either this horizontal Fn or the grounds Ff be 0 since one of them is the pivot point?
     
  15. Apr 17, 2006 #14

    arildno

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    No.
    Forces acting on the arbitrarily chosen "pivot point" produce no torque about THAT point, but the force need not be zero. It is the TORQUE that vanishes, because the distance is zero, not that the force is zero!
     
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