Solve Tension & Torque for Max Distance x

[Destrio]

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)

 

[Destrio]

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

 

[ogb p]

and the maximum tension of 520N.

Yes, your approach so far is correct. To solve for x, you can use the fact that the wire will break when the tension reaches its maximum value of 520N. So, you can set T = 520N in your equation T = Wx/L and solve for x.

520N = (Wx)/L
x = (520N*L)/W

Now, plug in the given values for W, L, and theta to find the maximum distance x before the wire breaks:

x = (520N*2.76m)/(315N) = 4.57m

Therefore, the maximum distance x before the wire breaks is 4.57m.