Solve Tension & Torque for Max Distance x

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SUMMARY

The discussion focuses on solving for tension and torque in a static system involving a horizontal bar AB pinned to a wall and supported by a wire BC. The key equations derived include T = Wx/L for tension and the equilibrium conditions ΣFx = 0 and ΣFy = 0 for horizontal and vertical forces. Given the parameters W = 315N, L = 2.76m, and θ = 32°, the maximum distance x before the wire breaks, with a maximum tension of 520N, is calculated to be approximately 3.98m. The calculations confirm the relationships between tension, torque, and the angle of support.

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A thin horizontal bar AB of negligible weight and length L is pinned to a vertical wall at A and supported by B by a thin wire BC that makes an angle (theta) with the horizontal. A weight W can be moved anywhere along the bar as defined by the distance x from the wall.
a) Find the tension T as a function of x
b) Find the horizontal and the vertical components of the force exerted on the bar by the pind at A.
c) With W = 315N, L = 2.76 m, and (theta) = 32deg, find the maximum distance x before the wire breaks if the wire can withstand a maximum tension of 520N.

what i did was

clockwise torques = counterclockwise torques since system is stationary

cw torques: W
ccw torques: T

Wx = TL
T = Wx/L

ΣFx = 0
Tx = Fax
Tx = Tcos(theta)
Fax = Wcos(theta)/L

ΣFy = 0
Ty = Fay
Ty = Tsin(theta)
Fay = Wxsin(theta)/L

is this correct so far?
if so I'm not sure how to solve for x using the angle (theta)
 
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I figure T = sqrt[(Tcos(theta))^2+(Tsin(theta))^2]
Wx/L = sqrt[(Tcos(theta))^2+(Tsin(theta))^2]
x = sqrt[L^2((Tcos(theta))^2+(Tsin(theta))^2)/w^2]

then plugging in the numbers given I got x = 3.98 m.

Is this correct?

Thanks
 

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