Question

End of A of a uniform rod AB of weight W and length a is hinged freely to a vertical wall B is connected to C on the wall by a light inextensible string of length a c is at b distant above A show the reaction of hinge at A is w/4b √[a²+8b²]

Answer 1

Solution:

About the hinge, the torque due to the tension in the string is equal to the torque due to the weight:

T*sin(2Θ)*a = W*($\tfrac{a}{2}$)*cosΘ

substitute sin(2Θ) = 2sinΘcosΘ and cancel the a*cosΘ terms:

T*2*sinΘ = $\dfrac{W}{2}$

T = $\dfrac{W}{4sin\theta}$

But sinΘ = $\tfrac{b}{2a}$ , so

T = $\dfrac{W}{4(\frac{b}{2a})}=\dfrac{aW}{2b}$

which has vertical component

Ty = T*sinΘ = $(\dfrac{aW}{2b})*(\dfrac{b}{2a})=\dfrac{W}{4}$

and horizontal component

Tx = T*cosΘ = $(\dfrac{aW}{2b})$*$\dfrac{√(a² - b²/4)}{a}$ = $(\dfrac{W}{2b})$ *$√(a² - b²/4)$

That makes the reaction components at the hinge

Rx = Tx

and Ry = W - $\dfrac{W}{4}$ = $\dfrac{3W}{4}$.

Then

R = √[$(\dfrac{W}{2b})$²*(a² - $\frac{b^2}{4}$) + ($\frac{3W}{4}$)²]

R = $(\dfrac{W}{2b})$ *√[a² - $\frac{b^2}{4}$ + $\frac{9b^2}{4}$ ]

R = $(\dfrac{W}{2b})$ *√[a² + 2b²] ◄

I have been over this several ways and believe that the "target" you provide is wrong.