## College Physics (4th Edition)

The tension in the wire that makes an angle of $15^{\circ}$ with the vertical is $29.6~N$ The tension in the wire that makes an angle of $25^{\circ}$ with the vertical is $18.1~N$
The horizontal component of the tension in each wire is equal in magnitude: $T_1~sin~15^{\circ} = T_2~sin~25^{\circ}$ $T_1 = \frac{T_2~sin~25^{\circ}}{sin~15^{\circ}}$ The sum of the vertical component in each wire is equal in magnitude to the lithograph's weight: $T_1~cos~15^{\circ} + T_2~cos~25^{\circ} = 45~N$ $(\frac{T_2~sin~25^{\circ}}{sin~15^{\circ}})~cos~15^{\circ} + T_2~cos~25^{\circ} = 45~N$ $T_2~(sin~25^{\circ}cot~15^{\circ} + cos~25^{\circ}) = 45~N$ $T_2 = \frac{45~N}{(sin~25^{\circ}cot~15^{\circ} + cos~25^{\circ})}$ $T_2 = 18.1~N$ We can find $T_1$: $T_1 = \frac{T_2~sin~25^{\circ}}{sin~15^{\circ}}$ $T_1 = \frac{(18.1~N)~sin~25^{\circ}}{sin~15^{\circ}}$ $T_1 = 29.6~N$