Solution
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To convert a length from meters to feet, we need to understand the conversion factor between these two units. The conversion factor from meters to feet is approximately 3.28084. This means that one meter is equivalent to 3.28084 feet.
Step 1: Identify the conversion factor
Step 1: Identify the conversion factor
Step 1: Identify the conversion factor
The conversion factor from meters to feet is 3.28084. This means that one meter is equivalent to 3.28084 feet.
Step 2: Multiply the number of meters by the conversion factor
To convert 20 meters to feet, we multiply 20 by the conversion factor of 3.28084.
20 meters * 3.28084 = 65.6168 feet
So, 20 meters is approximately 65.62 feet when rounded to two decimal places.
In conclusion, when you convert 20 meters to feet using the conversion factor of 3.28084, you find that 20 meters is approximately 65.62 feet. This conversion is useful in various fields, including science, engineering, and everyday life, as it allows for a better understanding of measurements in different units.
The conversion factor from meters to feet is 3.28084. This means that one meter is equivalent to 3.28084 feet.
Step 2: Multiply the number of meters by the conversion factor
To convert 20 meters to feet, we multiply 20 by the conversion factor of 3.28084.
20 meters * 3.28084 = 65.6168 feet
So, 20 meters is approximately 65.62 feet when rounded to two decimal places.
In conclusion, when you convert 20 meters to feet using the conversion factor of 3.28084, you find that 20 meters is approximately 65.62 feet. This conversion is useful in various fields, including science, engineering, and everyday life, as it allows for a better understanding of measurements in different units.
Man On A Boat
Exercise 2 - Conversions of units (mass, length, surface)
Exercise 3 - Conversions of units (length, mass, volume)
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Man On A Boat
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Hello, in this question,
we're told that a man is standing at one end of a boat,
and that the boat is of length 3 meters.
We're told that the man weighs 70 kilograms and that the boat weighs 100 kilograms.
The man moves then 2 meters up the boat.
He's starting at one end and he walks until here.
How much does the boat move?
Then we're being told that we can ignore the friction between the boat and the water.
Let's just speak about why the boat moves when the man is walking.
In order for the man to move,
there must be some friction between his feet and the boat,
which is allowing him to push himself forward.
The friction that is acting over here is static friction.
His foot lands on the floor of the boat.
This is his foot on the floor of the boat and as he lifts
his foot he's pushing
the floor of the boat backwards slightly.
That's him pushing forward.
As we know, the floor is moving back slightly because of Newton's third law
saying that there's an equal and opposite force to every force.
That means that here,
there's also static friction.
Now, if we look at these 2 forces and then we try and work out this question,
it's going to be very tricky and very difficult.
An easier way to do this is by using,
our now new found trick,
which is by using the center of mass.
Watch how easy this is.
Let's see, first of all,
we can rub this out because we don't need this.
We already know that the man weighs 70 kilograms and the boat weighs 100 kilograms.
Then we can say that the center of mass is around about,
let's say it's over here.
Soon we'll work it out formally,
but let's say it's around about here.
Now we can say that all the friction that's happening
between the man walking on the boat is now an internal force.
As we know in dealing with center of mass,
it's only the external forces which matter.
The friction between the man in the boat is internal,
so we can just ignore it and because we're
being told that we can ignore the friction between the boat and the water.
That means that we have no external forces.
Now, what does this mean if there are no external forces?
Also the friction between the man and the boat is internal,
so not counted. I wrote down these points.
That means that we can use this idea that because there's no external forces,
it means that the center of mass must remain constant.
It means that the center of mass is not moving and therefore,
when the man is standing at one edge of the boat,
and when the man is standing 2 meters in the boat,
the center of mass, must have stayed in the same place.
What I'm going to do is I'm going to work out
the center of mass of the man at the beginning,
the center of mass of the man at the end and then
I'm going to equate them one to the other.
It means that the center of mass is conserved.
Then we can find how much the boat has moved through that.
Let's see this in action.
Now let's start by finding the center of mass at the first moment.
First I have to find where I'm going to put
my axes and from there I work out the center of mass.
I'm going to say that my axes is right over
here and we're going to work this out according to our x center of mass.
Now of course, we can also do our y center of mass,
but here it doesn't matter because we can see that the boat is only moving in the x axis.
As we know, we have the sum of all of the masses multiplied by their possessions,
divided by the sum of all of the masses.
We'll put in 1 over here.
At the beginning, the man,
which is 70 kilograms,
is located 3 meters away,
so multiplied by 3 meters plus the boat which weighs 100 kilograms.
Then we're going to multiply it by,
what are we multiplying it by?
Because we can see clearly that the boat isn't a point-mass.
It's some long body.
What we have to do is we have to find its center of mass.
Now in the question, we don't know if the boat is homogeneous,
we don't know if one end is heavier than the other,
so it doesn't matter.
You'll see in just a moment.
Let's say that its center of mass is over here at a distance d away from the origin.
I'll just multiply the center of mass by the mass of the boat.
Then we divide it by the total mass,
which as we can see, is 170 kilograms.
This is the center of mass right at the start.
Now in front of us is the position of the boat after the man has walked these 2 meters.
Now notice the boat has moved.
Our axes that we started off with is in the exact same position.
Exactly the same place, and the boat has just moved away from it.
Let's say that we're going to call this distance between the axis and the boat,
let's say that this is a distance x.
Now, our d which was
the distance from the axis until the center of mass is still going to be the same.
We can draw it over here.
This is our d it's exactly the same.
Our boat's length is still exactly the same 3 meters,
and our guy now is standing at 2 meters into the boat.
Now let's work out what the new center of mass is. We have xcm_2.
Again, that's the sum of the masses multiplied by
their positions divided by the total mass of the system.
Now we're going to look at the position of the man.
The man is at distance x away from the axes plus 1 meter, because he's standing.
This distance is 1 meter and then this distance is 2 meters,
because the boat is 3 meters long.
We have an x plus 1 meter,
so x plus 1 meter multiplied by his mass,
which is 70 kilograms.
Now notice also that the center of mass is in the exact same position.
It also hasn't moved because that is the basis of this calculation.
We're saying that the center of mass has not moved.
You can see it in the picture here and here it's at the exact same position.
Let me just clear this.
Now we're going to do this for the boat as well.
Now, as we know with the boat,
we have to find the center of mass of the boat just like before.
The center of mass, again,
is a distance d from one end of the boat until the center of mass plus the x.
The center of mass of the boat, which is here,
is now a distance of d plus x away from the origin of the axes.
Plus, and then we're going to have x plus d multiplied by the mass of the boat,
which is 100 kilograms and then divided by the total mass of the system,
which is 170 kilograms.
Now what I have over here in xcm_2 has to equal what I have in xcm_1,
because that means that the center of mass has not moved.
Let's equate both sides.
We can see immediately that the denominator,
we don't have to write it because we can just multiply
both sides by 170 kilograms and get rid of it.
Then we'll see that we have 70 times 3.
We can not look at the units in the meantime.
We'll have 210 plus
100d is going to be equal to 70x I'm opening up the brackets here,
plus 70 plus 100x plus 100d.
Then we see that we have a 100d on both sides so we can subtract from both sides 100d.
That's why at the beginning when I said what this d is,
it doesn't really matter because at the end it's going to cancel out.
Which is a really good thing because if you're trying to
find in real life the center of mass of a boat,
it could be quite tricky.
Here it cancels out.
Then we can subtract 70 from each side,
so we'll have a 140 is equal to 170x.
Then in order to isolate out the x and the x is in fact what we were looking for,
it's how much the boats moved.
It's just going to be equal to 14 divided by 17 meters.
Meters, are the units,
I'll just symbolize it like that.
That means that when the man is standing on one end of
the boat and moves 2 meters down the boat,
the boat moves almost 1 meter,
not 1 meter a bit less this distance.
That's how to solve a relatively complicated question using this idea of center of mass.
Also, when answering this type of question,
mostly you'll be told that the boat moves and you
just have to know that the center of mass remains in the exact same position.
That's how we answer this type of question,
but we have to make sure that the center of mass starts at rest.
Before there's any movement the boat is stationary and the center of mass is stationary.
If the boat is moving at the beginning,
then we can't solve it like this.
We're going to have to use a combination of kinematics and the center of mass.
But that's less common in these types of exam questions.
We have to make sure that the center of mass starts at rest,
and very important that there are no external forces acting.
If they are external force is acting,
then it's a different type of question.
Now, also note that here,
we were asking the position where the boat will be.
How much it will move, where the boat will be.
We're being asked about the position.
If however, we were asked about something else,
the motion of the boat,
the velocity with which the boat would be moving,
then we would be looking at a different way of solving this type of question.
Not necessarily using the center of mass,
but rather using maybe conservation of energy or momentum and things like that.
This video explains how to calculate the displacement of a boat when a man of 70 kg moves 2 meters up the boat, which is 3 meters long and weighs 100 kg. The trick is to use the concept of the center of mass, which remains constant when there are no external forces acting. The center of mass is calculated by summing the masses multiplied by their positions, divided by the total mass. The boat moves 14/17 meters when the man moves 2 meters.
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