Answer to Question #158575 in Mechanics | Relativity for KHUZAIMA

Question

a uniform rod AB of length a and weight w is freely hinged to a vertical wall at A and is maintained in equlibrium by a light string of lenght a fastened to B and to a point C at a distance b vertically above A. prove reaction at hinge WSqrt(a²-2b²)/2b and find tension in the string

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(2$\theta$ )*a = W*($\tfrac{a}{2}$)*cos$\theta$

substitute sin(2$\theta$ ) = 2sinΘcosΘ and cancel the a*cos$\theta$ terms:

T*2*sin$\theta$ = $\dfrac{W}{2}$

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

But sin$\theta$ = $\tfrac{b}{2a}$ , 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 = $(\dfrac{aW}{2b})$*$\dfrac{ \sqrt{\smash[b]{(a² - \frac{b^2}{4})}}}{a}$ = $(\dfrac{W}{2b})$ *$\sqrt{\smash[b]{(a² - \frac{b^2}{4})}}$

That makes the reaction components at the hinge

Rx = Tx

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

Then

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

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

R = $(\dfrac{W}{2b})$ *$\sqrt{\smash[b]{[a² + 2b²]}}$ ◄