(a) [tex]\vec{F} = 1.0 \vec{i} - 2.0 \vec{j}[/tex]
The two forces are:
[tex]\vec{F_1} = 3.0 \vec{i} + 4.0 \vec{j}\\\vec{F_2} = -2.0 \vec{i} -6.0 \vec{j}[/tex]
To calculate the net force, we must calculate the resultant of the two vectors, which is obtained by separately summing the components of each vector:
[tex]F_x = 3.0 \vec{i}+(-2.0) \vec{i} = 1.0 \vec{i}\\F_y = 4.0 \vec{j} + (-6.0) \vec{j} = -2.0 \vec{j}[/tex]
So, the net force is
[tex]\vec{F} = 1.0 \vec{i} - 2.0 \vec{j}[/tex]
(b) 2.24 N
The magnitude of the net force is given by the Pythagorean theorem: it is equal to the square root of the sum of the squares of the single components:
[tex]|F|= \sqrt{F_x^2+F_y^2}=\sqrt{(1.0)^2+(-2.0)^2}=\sqrt{5}=2.24 N[/tex]
(c) [tex]-63.4^{\circ}[/tex]
The angle of the net force, relative to the positive x direction, is equal to the arctangent of the ratio between the vertical component and the horizontal component:
[tex]\theta=tan^{-1} (\frac{F_y}{F_x})=tan^{-1}(\frac{-2.0}{1.0})=-63.4^{\circ}[/tex]
(d) 2.24 m/s^2
The mass of the object is m = 1 kg, so according to Newton's second law the acceleration is given by
[tex]\vec{a} = \frac{\vec{F}}{m}[/tex]
Since we are interested in the magnitude of the acceleration, we have to take the magnitude of the net force into the calculation:
[tex]|a| = \frac{|F|}{m}=\frac{2.23 N}{1.0 kg}=2.24 m/s^2[/tex]
(e) [tex]-63.4^{\circ}[/tex]
According to Newton's second law:
[tex]\vec{a} = \frac{\vec{F}}{m}[/tex]
The acceleration has the same direction of the force, therefore the angle of the acceleration (measured with respect to the positive x direction) is [tex]-63.4^{\circ}[/tex].
