00:01
Students, in this problem we have to find the vector sum of these vectors according to the component method so the magnitude of vector a is 2 .25 centimeter and magnitude of vector b is 6 .35 centimeter and magnitude of vector c is 5 .47 centimeter and magnitude of vector d is 4 .19 centimeter so first of all we will write the vector a in magnitude form sorry in component form so this will be equal to this is the x component that will be equal to 2 .25 cost 60 i gap plus 2 .25 5 5 xx now vector b will be equal to 6 .35 cost 20 i cap this is x component of vector b and this will be the y component here so this is towards negative y axis so this will be minus 6 .35 sine 20 j cap now vector c will have two component so the x component will be this and y component will be this so the x component of vector c will be 5 .47 sine 36 i cap and y component will be 5 .47 cost 36 j cap now vector d will be equal to this is vector d so this will be x component towards negative x -axis and this will be y component towards positive y axis.
02:06
So this will be 4 .19 cost 36 in negative plus 4 .19 sine 36 jkamp.
02:23
So these are the component form of vector a, v, c and d.
02:28
So we have to find the resultant of the four displacement shown in the figure.
02:36
So the resultant vector will be.
02:38
Vector a plus vector b plus vector c plus vector t so we will add the x component together these these are the x component so we can write the resultant vector will be 2 .25 cost 60 i cap cost 60 plus 6 .35 cost 20 plus 5 .47 sign 36 and minus 4 .19 cost 36 this is in icap direction now we will add y component so this will be 2 .25 sine 60 6 .35 sine 20 in negative and negative 5 .47 cost 36 and this is 4 .191929 this is in j cap direction.
03:59
So this is the resultant vector.
04:00
We will calculate this.
04:03
This is equal to.
04:06
So this is the resultant vector.
04:09
X component is 6 .92 centimeter and y component is 2 .19 centimeter.
04:16
Now in part b, we have given vectors a, b, c and it is given that the vector sum is 0.
04:26
So we have to find the vector d and angle theta so we have to find this angle theta now in this case we have given the magnitude of vector a is 3 centimeter and magnitude of vector b is 7 cm and the magnitude of vector c is given which is 6 centimeter so the sum of all vector is equal to 0 so we will write vector a in vector form this is 3 cost 60 i k plus 3 sine 60 jkv right so this will be 3 into cost 60 that will be equal to 1 .5 i cap plus 3 into sin 60 so if we calculate this this is 2 .6 jkk now again vector b will be equal to 7 cost 20 i cap and the y vector will y component will be in the negative y axis towards the negative y axis so this will be minus 7 sine 20 j cap again we will calculate this 7 into pos 20 that is equal to 6 .58 i cap and this is 7 into sine 20 this will be equal to 39 j cap now vector c will be equal to the x component will be this in the y component will be this x component will be 6 sine 36 i cap minus 6 cost 36 j cap so this is 6 into sign 36 there is 3 .53 i cap minus 6 into cost 36 that is equal to 4 .85 j -cath.
06:55
Now we have given that vector a plus vector b plus vector c plus vector d is equal to 0.
07:04
Suppose vector d has component d x i cap plus d -y -j -cap.
07:14
Suppose we know, we don't know but x component is d x and y component is d -y.
07:19
Now we will add all these vectors.
07:22
So vector a is 1 .5 plus 2 .6.
07:27
So this will be 1 .5 icap.
07:31
Now we will add x component together.
07:34
So the x component of vector v is 6 .58, x component of vector c is 3 .53 and x component of vector d is d x.
07:48
This is in i cap...