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
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)
Thus,
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
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
x=L/ (tanφ₁*cotφ₂)+1