First, separate each vector into its vertical and horizontal components. Then, add all the vertical components together, as well as the horizontal components, to find the components of the vectpr F+T+N. Set each component of the vector F+T+N to be equal to the corresponding component of 2.00A to find the components of A. Use the components of A to find the magnitude and direction of the vector A.
The vertical and horizontal components of T are given by:
[tex]\begin{gathered} T_y=T\sin (10.0º) \\ =75N\cdot0.1736\ldots \\ =13.02N \end{gathered}[/tex][tex]\begin{gathered} T_x=T\cos (10.0) \\ =75N\cdot0.9848\ldots \\ =73.86N \end{gathered}[/tex]The vertical and horizontal components of N are:
[tex]\begin{gathered} N_y=N \\ =25.6N \end{gathered}[/tex][tex]N_x=0[/tex]The vertical and horizontal components of F are:
[tex]\begin{gathered} F_y=-F\cdot\cos (15) \\ =-40N\cdot0.9659\ldots \\ =-38.637N \end{gathered}[/tex][tex]\begin{gathered} F_x=-F\cdot\sin (15) \\ =-40N\cdot0.2588\ldots \\ =-10.353N \end{gathered}[/tex]The vertical component of F+N+T is given by:
[tex]\begin{gathered} (F+N+T)_y=F_y+N_y+T_y \\ =-38.637N+25.6N+13.02N \\ =-0.017N \\ \approx0.0N \end{gathered}[/tex]The horizontal component of F+N+T is given by:
[tex]\begin{gathered} (F+N+T)_x=F_x+N_x+T_x \\ =-10.353N+73.86N \\ =63.51N \end{gathered}[/tex]Since F+N+T=2.00A, then this is true for each component:
[tex]\begin{gathered} (F+N+T)_y=(2.00A)_y=2.00\cdot A_y \\ (F+N+T)_x=(2.00A)_x=2.00\cdot A_x \end{gathered}[/tex]Substitute the values for the components of F+N+T to find the values of the components of A:
[tex]\begin{gathered} 2.00A_y=0.00N \\ \Rightarrow A_y=\frac{0.0N}{2.00}=0.0N \end{gathered}[/tex][tex]\begin{gathered} 2.00A_x=63.51N \\ \Rightarrow A_x=\frac{63.51N}{2.00}=31.75N \end{gathered}[/tex]Since the vertical component of A is 0, then the vector A has a magnitude of 31.75N directed toward the positive X axis.
Therefeore, the answer is:
[tex]A=31.75N\text{ toward the positive direction of X.}[/tex]