Answer:
[tex]T_1=409.95\ N\\T_2=502.08\ N[/tex]
Explanation:
Conditions for Equilibrium
There are three forces applied to Bethany: T1 to the right making an angle of 30° with the ground, T2 to the left making an angle of 45° with the ground, and the weight W down the y-axis. Since the system is assumed to be at rest, the forces must be in equilibrium.
Please refer to the image below.
In the y-axis:
[tex]T_{1y}+T_{2y}=W[/tex]
The projections of the tensions in the vertical direction are
[tex]T_{1y}=T_1sin30^o[/tex]
[tex]T_{2y}=T_2sin45^o[/tex]
Thus, replacing into the first equation:
[tex]T_1sin30^o+T_2sin45^o=W\text{...........[1]}[/tex]
Now for the x-axis
[tex]T_{1x}-T_{2x}=0[/tex]
[tex]T_{1x}=T_{2x}[/tex]
The components of T1 and T2 in the horizontal direction are
[tex]T_{1x}=T_1cos30^o[/tex]
[tex]T_{2x}=T_2cos45^o[/tex]
Replacing
[tex]T_1cos30^o=T_2cos45^o[/tex]
Solving for T1
[tex]\displaystyle T_1=\frac{T_2cos45^o}{cos30^o}[/tex]
[tex]T_1=0.8165T_2[/tex]
Replacing in [1]
[tex]0.8165T_2sin30^o+T_2sin45^o=560[/tex]
Computing the trigonometric values and simplifying
[tex]1.1154T_2=560[/tex]
Solving
[tex]\boxed{T_2=502.08\ N}[/tex]
And therefore
[tex]T_1=0.8165T_2=0.8165\cdot 502.08\ N[/tex]
[tex]\boxed{T_1=409.95\ N}[/tex]
