Question

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 force F1 of magnitude 5.90 units acts on a crate at the origin in a direction = 26.0° above the positive x-axis. A second force F2 of magnitude 5.00 units acts on the crate in the direction of the positive y-axis. Find graphically the magnitude and direction (in degrees counterclockwise from the +x-axis) of the resultant force F1 + F2.

Answer 1

First, let's draw the resultant vector. To add the vectors, let's draw the start point of one vector in the endpoint of the other vector. Then, the resultant vector will have the start point of the first vector and the endpoint of the second vector:

To find the magnitude and direction of the resultant vector, let's decompose F1 in horizontal and vertical directions:

`\begin{gathered} F_(1x)=F_1\cos26°\\ \\ F_(1x)=5.9\cdot0.8988\\ \\ F_(1x)=5.3\\ \\ \\ \\ F_(1y)=F_1\sin26°\\ \\ F_(1y)=5.9\cdot0.438\\ \\ F_(1y)=2.58 \end{gathered}`

Since F2 is in the vertical direction, we have F2x = 0 and F2y = F2 = 5.

Now, calculating the magnitude and direction of the resultant vector:

`\begin{gathered} F_(rx)=F_(1x)+F_(2x)=5.3+0=5.3\\ \\ F_(ry)=F_(1y)+F_(2y)=2.58+5=7.58\\ \\ \\ \\ F_r=\sqrt{F_(rx)^2+F_(ry)^2}\\ \\ F_r=√(5.3^2+7.58^2)\\ \\ F_r=9.25\\ \\ \\ \\ \theta=\tan^(-1)((F_(ry))/(F_(rx)))\\ \\ \theta=\tan^(-1)((7.58)/(5.3))\\ \\ \theta=55.03° \end{gathered}`

Therefore the magnitude is 9.25 units and the direction is 55.03°.