A Bar Suspended by Two Wires?

A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle phi_1 with the horizontal, and the right wire makes an angle phi_2. The bar has length L. (Intro 1 figure)

<img src=”http://session.masteringphysics.com/problemAsset/1…

Find the position of the center of mass of the bar, x, measured from the bar’s left end.

Express the center of mass in terms of L, phi_1, and phi_2.

I typed in x = L / ( (tan(φ₂)/tan(φ₁) + 1), but its telling me that its incorrect, what do I do?

3 Answers

  • YT
    1 month ago

    Your answer, I think, is incorrect.

    -1st, you have to write the equation for Torques about left end of bar:

    L*T_2*sin(phi_2) – mgx = 0

    – Equation for the x-components of the forces:

    T_2*cos(phi_2) – T_1*cos(phi_1) = 0

    ==> T-1/T-2 = cos(phi-2)/cos(phi-1)

    – Equation for the y-components of the forces:

    T-1*sin(phi-1) + T-2*sin(phi-2) – mg = 0

    ==> mg = T-1*sin(phi-1) + T-2*sin(phi-2)

    – Find x from the 1st equation. To eliminate mg from the 1st equation, replace mg by T-1*sin(phi-1) + T-2*sin(phi-2)


    x = T-2*sin(phi-2)*L / (T-1*sin(phi-1) + T-2*sin(phi-2))

    Divide top and bottom by T-2*sin(phi-2), you have:

    x = L / (T-1*sin(phi-1) / T-2*sin(phi-2) + 1)

    From 2nd equation, you have T-1/T-2 = cos(phi-2)/cos(phi-1), replace this to the above x equation to eliminate T-1/T-2

  • James
    6 days ago

    The answer is x = L tan(phi_2)/tan(phi_2)+tan(phi_1)

    don’t forget the parenthesis – causes an error…lol took 19 tries

  • A S
    1 month ago

    x=L/ (tanφ₁*cotφ₂)+1

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