00:01
In this problem, we can use newton's second law of motion and what we know about vector addition to find the answers.
00:08
First of all, let's find the net force for part a.
00:12
To do that, we need to add these two forces, but since they are already fully on the x and y -axis, we don't need to decompose them into individual components.
00:22
So we'll simply write that the net force is equal to the square root of force 1 squared plus force 2 squared and we're given that force 1 is 10 .2 newtons and force 2 is equal to 16 newtons and we get that this is roughly 19 newtons and this is a magnitude that may be in between here but we don't know exactly where it is and to find that exact point we need to find the angle and to do that, we'll take the inverse 10 of f1 over f2.
01:29
And when we do that, we obtain an angle of roughly 32 .5 degrees.
01:49
This angle is going to be from the y -axis, from the negative y -axis, up to that vector.
01:55
So let's say that our net force is somewhere over here.
02:05
The angle we just found is this one.
02:07
So here we have our angle and here we have our net force.
02:20
And these are our final answers.
02:24
But we also need to find the acceleration.
02:27
And to do that, we can use newton's second law of motion, which tells us that the net force is equal to the mass times the acceleration.
02:39
Therefore, the acceleration will be equal to the net force over the mass.
02:49
And we have up here the magnitude of the net force and we're given the mass to be 18 .5 kilograms and that will give us roughly 1 meter per second squared and if we were to write this as a vector remember that it would be parallel to the net force because net force and the acceleration are always parallel.
03:29
There we go.
03:31
Now let's move on to part b.
03:36
In part b we can follow the same logic, but this time we have a vector that is not fully on the y or x -axis.
03:44
It has components on both of them.
03:47
So we are going to have to decompose it.
03:52
So let's begin talking about the y -axis.
04:00
The forces the y -axis will be equal of course to f2, which is fully on that axis and then we have this component that is in the negative portion of the y axis so we'll have minus f1 and to find that component you're going to multiply f1 by sign of the angle now that angle is going to be this one over here so if we have 90 up here and we know that from this point up to this point there are 120 then this one angle must be 30 degrees.
04:46
So we'll take sine of 30 degrees.
04:51
That gives us 16 minus 10 .2 times sine of 30 and we get that our y component is going to be exactly 10 .9 newtons.
05:15
Let's do the same for the x -axis.
05:19
The x -axis, the only component we have, comes from f1...