Question

If the 1kg standard body is accelerated by only f1=(3.O)+(4.0) and f2=(-2.0)+(-6.0) then what is fnet(al in unit vector notation and as (b) magnitude and (c) an angle relative to the positive x direction? What are the (d) magnitude and (e) angle of a ?

Answer 1

(a) $\vec{F} = 1.0 \vec{i} - 2.0 \vec{j}$

The two forces are:

$\vec{F_1} = 3.0 \vec{i} + 4.0 \vec{j}\\\vec{F_2} = -2.0 \vec{i} -6.0 \vec{j}$

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:

$F_x = 3.0 \vec{i}+(-2.0) \vec{i} = 1.0 \vec{i}\\F_y = 4.0 \vec{j} + (-6.0) \vec{j} = -2.0 \vec{j}$

So, the net force is

$\vec{F} = 1.0 \vec{i} - 2.0 \vec{j}$

(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:

$|F|= \sqrt{F_x^2+F_y^2}=\sqrt{(1.0)^2+(-2.0)^2}=\sqrt{5}=2.24 N$

(c) $-63.4^{\circ}$

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:

$\theta=tan^{-1} (\frac{F_y}{F_x})=tan^{-1}(\frac{-2.0}{1.0})=-63.4^{\circ}$

(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

$\vec{a} = \frac{\vec{F}}{m}$

Since we are interested in the magnitude of the acceleration, we have to take the magnitude of the net force into the calculation:

$|a| = \frac{|F|}{m}=\frac{2.23 N}{1.0 kg}=2.24 m/s^2$

(e) $-63.4^{\circ}$

According to Newton's second law:

$\vec{a} = \frac{\vec{F}}{m}$

The acceleration has the same direction of the force, therefore the angle of the acceleration (measured with respect to the positive x direction) is $-63.4^{\circ}$.