Question

Three forces act on a moving object. One force has a magnitude of 83.7 N and is directed due north. Another has a magnitude of 59.5 N and is directed due west. What must be (a) the magnitude and (b) the direction of the third force, such that the object continues to move with a constant velocity? Express your answer as a positive angle south of east.

Answer 1

Answer:

• $|\vec{F}_3| = 102.92 \ N$

• $\theta = 57 \° 24 ' 48''$

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

$\vec{a} = \vec{0}$.

As the net force equals acceleration multiplied by mass , this must mean:

$\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}$.

So, the sum of the three forces must be zero:

$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}$,

this implies:

$\vec{F}_3 = - \vec{F}_1 - \vec{F}_2$.

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector $\hat{i}$ pointing east,  with the y axis unit vector $\hat{j}$ pointing south, so the positive angle is south of east. For this, we got for the first force:

$\vec{F}_1 = 83.7 \ N \ (-\hat{j})$,

as is pointing north, and for the second force:

$\vec{F}_2 = 59.9 \ N \ (-\hat{i})$,

as is pointing west.

Now, our third force will be:

$\vec{F}_3 = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})$

$\vec{F}_3 = 83.7 \ N \ \hat{j} + 59.9 \ N \ \hat{i}$

$\vec{F}_3 = (59.9 \ N , 83.7 \ N )$

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

$|\vec{R}| = \sqrt{R_x^2 + R_y^2}$

$|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}$

$|\vec{F}_3| = 102.92 \ N$

this is the magnitude.

To find the direction, we can use:

$\theta = arctan(\frac{F_{3_y}}{F_{3_x}})$

$\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })$

$\theta = 57 \° 24 ' 48''$

and this is the angle south of east.