物理一問，Help

For the given graph, first find the resultant force in the x and y directions
Resultant force in x direction = [2 + (-2)] N = 0 N
Resultant force in y direction = [1 + (-2)] N = -1 N
Hence, the resultant force is 1 N acting in the -y direction.

(a) Since the resultant force is 1 N acting in the -y direction, to keep the mass in equilibrium, the 3rd force must acts opposite to the resultant force.
Hence, the 3rd force = 1 N in the +y direction.

(b) The net force acting on the 1 kg mass = 1 x 2 N = 2 N in the +x direction.
Hence, the y-component of the applied force = 1 N
and the x-component of the applied force = 2 N
Magnitude of applied force = square-root[1^2 + 2^2] N = 2.236 N
and makes an angle of arc-tan[1/2] degrees = 26.6 degrees with the +x axis

(c) Since the mass is under retardation, the net force acts in the -y direction.
Net force (in the -y direction) = mass x acceleration = 1 x (-1) N = -1 N
Let F be the applied force acting on the mass
F + (-1) = -1
i.e. F = (-1+1) N = 0 N
Therefore, the applied force is zero.