Chapter 14.2, Problem 15E

To determine

The tension in the wire which is used to hang the mirror.

Explanation of Solution

Given information:

A $200kg$ load is hung with two cables. One wire makes an angle of $40°$ and another cable makes an angle of $60°$ with the horizontal.

Calculations:

Here we draw a schematic diagram of the situation.

  

Here point A to B is attached to the mirror and makes an angle of $30°$ with the horizontal.

Therefore gravitational force is acting downward.

Now we draw the force diagram

  

Now from the force diagram in point O,

  $\begin{array}{l}{F}_1\sin 40+F_2\sin 60=F_3\\ Also,\\ F_1\cos 40=F_2\cos 60\\ F_1=F_2\frac{\cos 60}{\cos 40}\\ F_1=0.65F_2\end{array}$

Now total force acting on the wire is,

  $F_3=200\times 9.8=1960N$

Therefore,

  $\begin{array}{l}{F}_1\sin 40+F_2\sin 60=1960\\ 0.65F_2\sin 40+F_2\sin 60=1960\\ 0.4168F_2+0.866F_2=1960\\ 1.212F_2=1960\\ F_2=1527N\\ Therefore,\\ F_1=0.65F_2=0.65\times 1527=993N\end{array}$

Therefore tension in AO is $F_1\sin 40=638N$

Tension in BO is $F_2\sin 60=1322N$