00:01
So here in this problem, the forces that are acting on the blocks are f1 force that's acting in this direction, and it has the magnitude of 4 newtons, and it's making an angle of 25 degree with respect to x -axis.
00:22
That's the positive x direction.
00:24
So here we need to consider the anti -clockwise direction as the positive direction.
00:32
And the other force that's acting on the block is f2, which is in this direction, and it is given that the f2 has magnitude of 6 newton, and this is making an angle of 325 degree with respect to positive x -axis.
00:57
And the third force is acting along the negative x direction, and that's having the magnitude as 8 newton.
01:10
So we need to figure out the resultant force that's acting on the block because of these three individual forces.
01:22
And here, in order to figure out the resultant force, we will be separately computing the magnitude of the resultant force and the angle that this resultant force makes with respect to positive x direction.
01:37
So let's first express all the forces in terms of icap and j cap.
01:43
So f1 we can see it will have two components, one along positive x direction, that is 4 cost 25 degree, and the other along positive y, that's 4 sine 25.
01:55
So let's write f1 in terms of vector.
01:57
So it will be 4 cost 25 degree i cap plus 4 sine 25 degree jcap.
02:05
So upon simplification, we get the value of 4 cost 25 as 3 .63.
02:16
So approximately 3 .63 eye cap.
02:19
And similarly, if we find the value of 4 times sine 25 degree, that comes out to be 1 .69 jcap.
02:29
So this is f1 vector and f3 vector we can say this is minus 8 i cap because it's along the negative x direction.
02:39
So here we have been given that 325 degree is the angle that f2 makes with positive x -axis and that 2 in the counterclockwise direction.
02:50
So here let's figure out this angle it will be 360 minus 325.
02:55
So 360 minus 325 that will be just 330 35 degree.
03:04
So now we can write f2 vector in terms of icap and j cap.
03:09
So it will come to some amount negative y direction.
03:16
So it will be 6 times cost 35 icap.
03:21
And the component along negative y, that will be minus 6 times sine 35 j cap.
03:30
All these sources are in standard units, that's the newton.
03:34
So when we multiply here cost 35 with 6 we get the result as 4 .9.
03:43
So this will be 4 .9 icap minus 6 times sine 35.
03:52
That comes out to be 3 .44 jcap.
03:56
So this is f2 vector.
03:59
So we have got f1 vector, f3 vector and f2 vector.
04:05
Figure out the resultant of this force the resultant vector will be the summation of these three forces vectorially so when we add them here we add the i components together so 3 .63 minus 8 plus 4 .9 we're going to do so that will come out to be 0 .53 .53 i cap similarly along j cap we will add them together so 1 .69 this has no j cap here so 1 .69 minus 3 .44 so that comes out to be minus 1 .75 a cap so this is the net force we have figured out and this is in the form of i cap and jcap together so now we need to find out the magnitude of this force so that magnitude will be given as root of 0 .53 square plus 1 .75 square.
05:18
So we first square them and then we add them together and finally we take the root that comes out to be 1 .83 approximately.
05:35
So 1 .83 newton will be the net magnitude of the resultant force.
05:41
So considering them in terms of two significant figures, the result will be 1 .8 newton.
05:50
So the magnitude will be 1 .8 newtons.
05:53
To get the direction, we need to consider the force that will be acting on the block.
06:00
So this is the net force.
06:02
That means on the block, one force is acting, which is 0 .53 newton to the right along x direction.
06:10
And another force that is 1 .75 newton...