Three vectors are shown in this figure. Their respective moduli are F = 40N T = 75 N and N = 25.6NCalculate the vector a so that F + N + T = 2.00A

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]