The tension is the same for each wire. The angle between each wire and vertical is 53˚. Let T be the tension. The vertical component of the tension in each wire is equal to one half the weight of the mirror.
Vertical component = T * cos 53
T * cos 53 = 40
T = 40 ÷ cos 53
This is approximately 66.5 N. If you go to the website below, you will see a drawing of a similar problem. The only difference is that the angles are measured from horizontal. I hope this is helpful for you.
Answers & Comments
The tension is the same for each wire. The angle between each wire and vertical is 53˚. Let T be the tension. The vertical component of the tension in each wire is equal to one half the weight of the mirror.
Vertical component = T * cos 53
T * cos 53 = 40
T = 40 ÷ cos 53
This is approximately 66.5 N. If you go to the website below, you will see a drawing of a similar problem. The only difference is that the angles are measured from horizontal. I hope this is helpful for you.
http://www.physicsphenomena.com/StudyofMotioninTwo...
angle between vertical and wire = ½ x 106° = 53°
Let tension in each wire = T
Now horizontal component of tension is balanced, while
vertical component of tension in both wires is counter balanced by the mirror s weight
2Tcos53 = 80
T = 40/cos53 = 66.47 N