Question 11 of 12 Current Attempt in Progress Three forces...

Question 11 of 12 Current Attempt in Progress Three forces are applied to an object, as shown in the figure. Force \(\vec{F_1}\) has a magnitude of 20.5 newtons (20.5 N) and is directed 30.0° to the left of the +y axis. Force \(\vec{F_2}\) has a magnitude of 15.4 N and points along the +x axis. What must be the (a) magnitude and (b) direction (specified by the angle \(\theta\) in the drawing) of the third force \(\vec{F_3}\) such that the vector sum of the three forces is 0 N?

Transcript

00:01 In this problem we are in the two -dimensional xy plane.

00:08 So this x this y and we have three forces f1, f2 and f3.

00:17 We know the magnitude and direction of f1 and also f2 but we don't know the magnitude or the direction of this third force f3.

00:32 But we know that it is such that this third force is such that the vector sum of all the force, all these three forces is zero.

00:44 So using this information, we are going to find the magnitude and the direction of this third force.

00:52 So let's place our forces.

00:58 Let's start with f1.

00:59 It says this f1.

01:03 Vector is directed 30 degrees to the left of the positive y axis so i'm going to call this direction theta so we have theta equal to 30 degrees and we are also given that the magnitude of these forces to not 221 .0 neutrons and now the second one this f2 force has a magnitude of 15 neutrons and it points along the positive x -axis so this is f2 now i'm going to put this unknown force in a very generic direction like that suppose i assume force in such a way that my angle measurement my direction is measured from the positive x axis towards the positive y axis namely in the counterclockwise direction so i'm going to call single phi okay now we have this complete picture now let's do some algebra we are given that this vector sum of all these three forces is zero namely we have the total x components equal to zero and also the total y components equal to zero let's write down each of these expressions so the total of all these x components let's start with f1 we have this component for the f1 it's a negative contribution due to the direction of this first force so we have minus sign magnitude of f1 times sine as for the second force we just include include it as is because it already points along the positive x -axis so we have f2 norm here.

03:49 Plus.

03:51 Okay, now this plus sign is important because i want to be consistent with my choice of direction.

03:58 I mean, phi can be negative at the end, but it doesn't matter.

04:03 As long as i know that i am measuring phi in this direction like this, it will be positive.

04:10 If it turns out to be negative, then it means i'm going in the other way.

04:14 That's okay.

04:16 So okay, i have this plus sign here.

04:20 The norm of this guy times cosine phi.

04:25 So this is equal to 0.

04:29 And let us solve this equation for cosine phi...

Question

Please give Ace some feedback