Question

Need Help? Red WRME My As you will see in a later chapter, forces are vector quantities, and the total force on an object is the vector sum of all forces acting on it. In the figure below, a free F1 of magnitude 6.70 units acts on a box at the gin in a direction θ-28.0° above the positive x-axis. A second force F2 of magnitude 5.00 units acts on the box in the of the positive y-axis. Find graphically the magnitude and direction (in degrees counterclockwise from the +x-axis) of the resultant force F1 + F2 magnitude units direction e counterclockwise from the +x-axis Need Help? Read I Submit Assignment Save Assignment Progres Home My Assignments Extension Reques copyright 02010 Cengege Leamng·inc Al Rights Resarved

Answer 1

In a two dimensional coordinate system, any vector can be broken into x-component and y-component.

In the figure shown below, the vector v is broken into two components,$v_x$ and $v_y$. Let the angle between the vector and its x-component be $\theta$.

The vector and its components form a right angled triangle as shown below.

The trigonometric ratios give the relation between the magnitude of the vector and the components of the vector.

$cos\,\,\theta=\frac{adjacent \,\, side} {hypotenuse}=\frac{v_x}{v}$

$sin\,\,\theta=\frac{opposite \,\, side} {hypotenuse}=\frac{v_y}{v}$

$\Rightarrow v_x=v\,\,cos\,\,\theta$

$v_y=v\,\,sin\,\,\theta$

Using the Pythagorean Theorem in the right triangle with length $v_x$ and $v_y$.

$|v|=\sqrt(v_x^2+v_y^2)$

Now looking into the given question, it is given that

$F_1=6.7\,\, units$ and $F_2=5\,\, units$

First, considering $F_1$ alone, angle between $F_1$ and the x axis is 28°.

Now,

$F_1__x=F_1\,\,cos\,\,\theta=6.7\,\,cos\,\,28=5.9$

  $F_1__y=F_1\,\,sin\,\,\theta=6.7\,\,sin\,\,28=3.1$

Now, considering the next force vector $F_2$ which is along the +ve y axis.

$\therefore F_2__x=0$ and

  $F_2__y=F_2=5$

Now, for the addition of two vectors, we simply add their components together.

Let the resultant force be F.

$\large F_x=F_1__x+F_2__x=5.9+0=5.9$

$\large F_y=F_1__y+F_2__y=3.1+5=8.1$

Now, F is given by,

$\large F=\sqrt(F_x^2+F_y^2)=\sqrt(5.9^2+8.1^2)=10$

We have,

  $\large F_x=F\,\, cos \,\,\theta$

$\large \therefore\,\, cos \,\,\theta=\frac{F_x}{F}=\frac{5.9}{10}=0.59$

  $\large \theta=cos^-^1\,\,0.59=53.8=54^o$

(You can also use the sin function to find the angle.)

Therefore, magnitude = 10 and

direction = 54° counter clockwise from the positive x-axis.