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Determining maximum force applied on a lever

  1. Jul 2, 2009 #1
    1. The problem statement, all variables and given/known data
    The lever BCD is hinged at C and is attached to a control rod at B. Determine the maximum force [tex]\vec{P}[/tex] which can be safely applied at D
    if the maximum allowable value of the reaction at C is 500 N.

    Here is a picture of the problem. It is number 4.20.
    "[URL [Broken] title="download file from Jumala Files"]http://jumalafiles.info/showfile2-14173678132920785343298714637103960/problem420.pdf [Broken][/URL]

    2. Relevant equations

    [tex]\Sigma F_{x} = 0[/tex]

    [tex]\Sigma F_{y} = 0[/tex]

    [tex]\Sigma M_{C} = 0[/tex]


    3. The attempt at a solution

    I know I have two unknowns at pin C [tex]C_{x}, C_{y}[/tex] but I don't know either of their magnitudes. I have a tension in rod BA and I don't its magnitude and I also don't know the magnitude of [tex]\vec{P}.[/tex] I only have 3 equations to work with here but more than that of unknowns. Am I missing a relationship here or am I not making a correct assumption?

    Thanks in advance,

    KEØM
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 3, 2009 #2

    nvn

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    But you also have a given equation regarding the magnitude of C. Therefore, don't you have four equations?
     
  4. Jul 3, 2009 #3
    I must just not be seeing how that equation helps me because I can't solve for any thing right now.

    [tex]\Sigma M_{B} = -(.03)C_{x} - (.04)C_{y} - (.105)P = 0[/tex]

    [tex]\Sigma F_{y} = C_{y} + Tsin(\theta) = 0[/tex]

    [tex]\Sigma F_{x} = C_{x} + P + Tcos(\theta) = 0[/tex]

    [tex] C = 500 = \sqrt{(C_{x})^2 + (C_{y})^2}[/tex]

    [tex]\theta = 53.1^{\circ}[/tex]
     
  5. Jul 3, 2009 #4

    nvn

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    Last edited by a moderator: Apr 24, 2017
  6. Jul 4, 2009 #5
    I have tried several times in solving these four equations and I just can't do it. Are you sure I am not missing some assumption or relationship?
     
  7. Jul 4, 2009 #6

    nvn

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    Your equations are correct. Use the advice in post 4. If you show your work, it will quickly uncover your algebra mistake.
     
  8. Jul 4, 2009 #7
    [tex]\Sigma M_{B} = -(.03)C_{x} - (.04)C_{y} - (.105)P = 0 [/tex]

    [tex]\Sigma F_{y} = C_{y} + Tsin(\theta) = 0[/tex]

    [tex]\Sigma F_{x} = C_{x} + P + Tcos(\theta) = 0[/tex]

    [tex]C = 500 = \sqrt{(C_{x})^2 + (C_{y})^2}[/tex]

    [tex]\theta = 53.1^{\circ}[/tex]

    [tex]C_{y} + Tsin(\theta) = 0 \Rightarrow C_{y} = -\frac{4}{5}T[/tex]

    [tex]C_{x} + P + Tcos(\theta) = 0 \Rightarrow C_{x} + P + \frac{3}{5}T = 0[/tex]

    [tex]T = -\frac{5}{4}C_{y}[/tex]

    [tex]C_{x} -\frac{1}{4}C_{y} = -P[/tex]

    [tex]-(.03)C_{x} - (.04)C_{y} - (.105)P = 0 \Rightarrow .2857C_{x} + 0.3810C_{y} = -P[/tex]

    [tex]C_{x} -\frac{1}{4}C_{y} = .2857C_{x} + 0.3810C_{y}[/tex]

    [tex]C_{x} -\frac{1}{4}C_{y} = 2857C_{x} + 0.3810C_{y}[/tex]

    [tex]0.7143C_{x} = 0.6310C_{y} \Rightarrow 1.13C_{x} = C_{y}[/tex]

    [tex]500^2 = C_{x}^2 + (1.13C_{x})^2[/tex]

    [tex]C_{x} = 331.02 N, C_{y} = 374.74 N[/tex]

    While I was typing my failed attempt in here I finally realized how to solve it.

    Is this right? When I try to solve for P I am getting a negative answer.

    Thanks again for helping me nvn,

    KEØM
     
  9. Jul 4, 2009 #8

    nvn

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    Close, but not quite; (3/5)*(5/4) is not equal to 1/4. Try it again. Also, always place a zero before the decimal point for numbers less than 1.
     
  10. Jul 4, 2009 #9
    Woops! Stupid Mistakes. Now I get:

    [tex]C_{x} = 422.7 N, C_{y} = 267.0 N[/tex]

    This doesn't change the negative value I get for P though.
     
  11. Jul 4, 2009 #10

    nvn

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    Very good, except notice that when you solve for Cx, it is actually Cx = +/-(number)^0.5. Knowing P is positive, you can figure out whether the other +/- values should be positive or negative.
     
  12. Jul 4, 2009 #11
    Ok. So if I choose [tex]C_{x}[/tex] to be negative then:

    [tex]\frac{3}{4}C_{y} - C_{x} = P \Rightarrow \frac{3}{4}(267.0) - (-422.7) = P[/tex]

    [tex] P = 623.0 N [/tex] and I have my answer

    Thanks again,

    KEØM
     
  13. Jul 4, 2009 #12

    nvn

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    No, plug your answer for Cx into an earlier equation to solve for Cy, to obtain the correct sign on Cy.
     
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