LESSON 2: Sum of Forces 2


Let’s continue to analyze this concept of summing forces by looking at the following example. This example is the same as past examples except in this example there is an angled force. In this example we will find the values of the two unlabeled vectors.


To begin this problem you should break the angled vector (blue) into its component parts. So, this 5 N vector is the same as having two vectors; one 4.33 N vector in the x-direction and one 2.5 N vector in the y-direction.

Now we can sum the forces. First let’s look at the y-direction: there are two downward forces, one having a magnitude of 2.5 N and the other 20N. 20 + 2.5 = 22.5 , so there is a total downward force of 22.5
N . There are also two upward forces, one having a magnitude of 7 N and the other is unknown. We know that the sum of the forces going down must equal the sum of the forces going up, so in order for the forces going up to equal 22.5 , the unknown force must equal 15.5 (22.5 - 7).

Now let’s sum the forces in the x-direction. In the x-direction there are two forces, one going to the right and one going to the left. In order for the bar to remain static, these two forces must be equal. We know the force to the right is 4.33 N, so the unknown force going to the left must also be 4.33 N.

So, by breaking the angled force into its component parts and then summing the forces in the y-direction and x-direction, we found the unknown vertical force to have a magnitude of 15.5 N and the unknown horizontal force to have a magnitude of 4.33 N.