# In the figure, one end of a uniform beam of weight 420 N is hinged to a wall; the other end is supported by a wire that makes angles θ = 29° with both wall and beam. ?

## How to find (a) the tension in the wire and the (b) horizontal and (c) vertical components of the force of the hinge on the beam. Feb 5, 2016

a)T=420 . cos 29=367.34 N
b)
$\text{horizontal component of net force :} 98 , 716 . \sin 2 \theta = 83 , 716 N$
c)
$\text{vertical component of net force :} 98 , 716 . \cos 2 \theta = 52 , 312 N$

#### Explanation: $P = 420 . \sin 2 \theta$
$R = 420 . \cos 2 \theta$
$L = T . \cos \theta$
$K = T . \sin \theta$
$M = T . \sin \theta$
$N = T . \cos \theta$
K .2.l=P. l(" torque for point A)"
$\text{ R and L have no torque for point A}$
$K .2 = P$
$2. T . \sin \theta = 420 . \sin 2. \theta$
$T = 210 . \frac{\sin 2 \theta}{\sin} \theta$
$\sin 2. \theta = 2. \sin \theta . \cos \theta$
$T = 210. \frac{2. \sin \theta . \cos \theta}{\sin} \theta$
$T = 420 . \cos \theta$
a)T=420 . cos 29=367.34 N
$L = 367.34 . \cos 29 = 321.282 N$
$R = 420 . \cos 58 = 222 , 566 N$
$\text{net force on point hinge :} 321 , 282 - 222 , 566 = 98 , 716 N$
$\text{horizontal component of net force :} 98 , 716 . \sin 2 \theta = 83 , 716 N$
$\text{vertical component of net force :} 98 , 716 . \cos 2 \theta = 52 , 312 N$