For simplicity, assume the banner and the bar have equal mass. please explain!!

Question

For simplicity, assume the banner and the bar have equal mass.

please explain!!

President Boren is hanging an OU banner from the outside of Boyd house. The banner is hanging from a bar of length L, and is to be supported by a cable that is attached a distance L/2 above the end of the bar. There are two options, to attach the cable to the end of the rod, or in the middle. For which of these choices is the tension in the cable the smallest? For simplicity, assume the banner and the bar have equal mass. please explain!!

Explanation / Answer

sum torque about attached end

- mg L + x T sin theta = 0

sin theta = L/2/(sqrt((L/2)^2 + x^2)

so

- mgL + x L/2 T/sqrt(L^2/4 + x^2) =0

x T/sqrt(L^2/4 + x^2) = mg

T = mg sqrt(L^2/4 + x^2)/x

when x = L

T = mg sqrt(L^2/4 + L^2)/L = mg *sqrt( 5/4)

when x=L/2

T = mg sqrt( 2 L^2/4)/L/2 = mg sqrt(2)

so less when atttached in the end

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---