00:01
There are five questions in this set and this is the first one.
00:04
A tight rope walker stands at the middle of a rope and the rope now makes an angle theta which is equal to 6 degrees with the horizontal.
00:15
The person here has a mass of 79 kilograms and we are to determine the tension along the wire.
00:24
So since we're talking about forces acting on the body here, a free body diagram will all be.
00:30
Be very useful.
00:31
So let's show the free body diagram of a point here on the rope where the person here stands on.
00:41
So as you can see, this portion of the rope, of course, is being pushed down by a force that's equal to the weight of the person.
00:50
And then the rope here, this portion will exert a tension in this direction, a pulling force.
00:57
And we also have another tension.
01:00
Here which which has equal magnitude but different different direction so we'll also call it as t that is what we are looking for so if we replace it i mean if we transfer the the forces on an xy plane we have here t and then another t okay and of course the force due to the weight of the tight rope walker and then this is theta.
01:33
This must also be theta.
01:35
Resolving the two components of tension, it has an x component to the right and a positive y component here, whereas the tension that is in the second quadrant will have negative tension along the x and a positive tension along the y.
01:51
The system here is an equilibrium.
01:54
So if we add all the forces in the y axis, it must be equal to zero because there's no acceleration in the y axis.
02:02
We don't want that to happen.
02:05
So adding algebraically, we have two positive forces here.
02:08
We have one ti -y and another ti -y.
02:12
So that makes it 2 -t -y.
02:15
And then you have weight going down, so that's a minus there.
02:19
Okay.
02:20
And then algebra takes over.
02:22
Ty -y is just resolved as t -t times sine.
02:27
And then we move weight to the right -hand side of the equation and weight now is equivalent to the mass times the magnitude of acceleration due to gravity so that the tension in the wire is just m g over 2 sine theta we are now ready to plug in our values okay so this is 79 kilograms acceleration due to gravity is 9 .8 and then we have 2 times sign 6 degrees.
03:02
We will declare our final answer with one significant figure because in the given we have 6 degrees as the number with the least number of significant figure which is 1.
03:18
So the calculator gives us 3 ,000, around 3 ,000.
03:24
7003 or 704.
03:28
But in our final answer, we'll declare it with one significant figure to be consistent.
03:33
So that is 1 ,000 newtons with one significant figure.
03:41
Okay.
03:42
So this is an example of a system in equilibrium.
03:47
Let's go to question number two.
03:48
In question number two, we are given here four forces with their components.
03:54
These forces are acting on a mass and the mass has six kilograms.
04:01
The question is to determine the resultant or the net force of the three.
04:06
So we will just add them vectorially.
04:12
So that would be the resultant force is just the vector sum of the three forces.
04:18
So we have f sub 2 plus f sub 3.
04:21
And then we will just put all together the components here.
04:27
So you just need to be very careful with the signs.
04:31
So this is our f1 plus our f2, negative 4i hat, plus 8j hat, plus k, 1k hat, and then plus the third vector, 5i, plus 2j minus 5k.
04:50
So algebraically, we can only add similar terms.
04:55
Physically, we can only add forces that are lying along the same axis.
05:00
So in this case, all terms here with unit vector i, this would be the x component.
05:09
So adding the 3 now, we have 2 plus negative 4 plus 5.
05:16
So that gives us positive 3.
05:18
Don't forget the unit vector.
05:22
And then negative 5, positive 8 j hat and positive 2 j hat.
05:28
These are all the forces along the y axis.
05:31
Adding them algebraically gives us positive 5 along the y axis.
05:37
And of course, the ones with the k hats are the forces along the zay axis.
05:44
And algebraically, it gives us negative 2 k hat.
05:48
Okay, there you go.
05:50
This is now the resultant force in terms of the components.
05:57
Okay, let's go to problem three.
05:59
Problem three, the situation is this.
06:02
You have a soccer ball with a mass of 0 .20 kilograms and you move it, you threw it to the right with unknown speed.
06:12
And then it collides with an antique with an antique vase of your mother.
06:20
The mass of the vase is 0 .80 kilograms and it's initially at rest.
06:26
So the ball hits the vase and then after collision, the ball moves to the opposite direction with a speed of 3 .9 meters per second.
06:36
So as a vector, we have your new sub 1 prime as the final.
06:42
Velocity of mass 1 and then you have the negative sign here whereas the vase moves to the opposite direction at 2 .6 meters per second so no sub 2 prime is the velocity of mass 2 after collision we are to determine the speed okay so take note how fast so that only refers to the speed of course speed is just the magnitude of the certain velocity.
07:13
So initially that would be v1.
07:15
Okay, so scalar quantity speed here, our answer will always be positive.
07:21
What speed at the initial state? okay, so if we neglect any friction here, we can see that momentum here is conserved and our ball -based system is isolated.
07:37
So for an isolated system, we know that the law of conservation of momentum says that whatever is the total momentum before collision, that will be equivalent to the total momentum of the system before collision.
07:53
So we'll just add the individual momentum here.
07:56
That would be the initial momentum of the ball, that would be m1, velocity 1, plus the initial momentum of the vase.
08:06
Okay.
08:07
But of course it's moving, it was a tressed initially.
08:11
So basically this is equal to zero.
08:14
And then final, that would still be mass 1.
08:17
Final velocity, so that's new 1 prime, plus the final momentum of the vase, nu sub 2 prime.
08:27
Okay.
08:28
So we are to determine the magnitude of no sub 1.
08:32
So we need to isolate it in the left hand side.
08:35
So that it becomes velocity 1 equals m1, no 1 prime, plus m2, no 2 prime, all over m1.
08:50
So that the first term here, we just have speed 1 final, plus the ratio of mass 2 to mass 1 times the final velocity of mass 2.
09:03
We're now ready to put in our values.
09:07
So we have here this is negative.
09:11
Do not forget that this is a vector quantity.
09:14
So it goes to the left.
09:16
There has to be negative sign along the x -axis.
09:19
And then you have here 0 .8 kilograms.
09:22
This one is just 0 .2 kilograms.
09:25
And this is positive 2 .6.
09:29
So plugging in the values, the calculator will give us.
09:36
6 .5 meters per second.
09:40
So as a vector quantity, this would be positive, and then i'll put here i -hat.
09:46
But the question is just how fast, it's just the speed.
09:50
So therefore, we get the absolute value here, which is still, of course, positive, but no more direction.
09:57
There's no more positive i -hat in our final answer.
10:00
So this is the initial velocity, i mean the initial speed of the ball in the situation here.
10:08
Okay, we have applied the conservation of momentum.
10:12
Let's go to number four.
10:15
We have here a solid cylinder that is rotating through a horizontal axis and the axis is just parallel to the cylinder as shown here.
10:24
There's a cable that is wrapped around it...