Video Transcript
An object has a weight of 45 newtons. What mass of water must the object displace to float? Answer to one decimal place.
Okay, so in this question, we’ve got an object. Let’s say this is our object. And we’ve been told that it has a weight of 45 newtons. Now, we’ve put this object in water and it’s floating. But in order for the object to float in the water, it has to displace some water in the first place.
Now in this situation, we can use something known as Archimedes’s principle. This principle tells us that the upthrust on an object when that object is placed in a fluid is equal to the weight of the displaced fluid. So in other words, this object that’s floating in water has an upthrust on it. We’ll call this upthrust 𝑇. And this upthrust is equal to the weight of the fluid that the object displaces.
Now, the displaced fluid is the fluid that was initially in this volume here. But that volume is now occupied by the floating object. And because we know the object is floating, the net force on the object must be zero. Because if the net force on the object wasn’t zero, then the object would accelerate in the direction of the overall net force. But it’s not doing that. The object is just floating. So the net force on the object is zero. Hence, the magnitude or size of this upthrust force here has to be the same as the magnitude of the weight of the object because these are the only two forces acting on the object and the forces must be balanced.
Now, the upthrust is acting in the opposite direction to the weight, which is 45 newtons. So if the upthrust is 45 newtons as well, then the two forces cancel each other out and the net force on the object is zero. Therefore, we realized that the upthrust is 45 newtons. But Archimedes’s principle tells us that this upthrust is equal to the weight of the displaced fluid — specifically the weight of the displaced water in this case. So we say that the weight of the displaced water is equal to 45 newtons.
Now, we’re getting close. We’re trying to find out the mass of the displaced water. So we need to find a relationship that links together the weight of the displaced water with its mass. The relationship that we’re looking for is that the weight of any object 𝑊 is given by multiplying the mass of that object by the gravitational field strength of the Earth 𝑔. Now, we can recall that the value of 𝑔 is 9.8 meters per second squared. That’s the acceleration due to gravity on Earth, also known as the gravitational field strength of the Earth.
And since we already know the weight of the displaced water, we can work out this mass. To do this, we need to rearrange a bit. We need to divide both sides of the equation by the gravitational field strength 𝑔. Doing so means that the 𝑔s cancel on the right-hand side, which leaves us with 𝑊 over 𝑔 is equal to 𝑚. Now at this point, we just need to plug in the values of the weight and the gravitational field strength. So the mass is going to be the weight which is 45 newtons divided by the gravitational field strength which is 9.8 meters per second squared.
And the good thing is that we’re using standard units. The weight is in standard unit of newtons and the gravitational field strength is in its standard unit of meters per second squared. Therefore, when we complete the calculation, the mass that we find is going to be in its standard unit of kilograms. So evaluating the right-hand side of the equation, we find that the mass of the displaced water is 4.591... kilograms.
However, we need to give our answer to one decimal place. So we need to round this value here. Now, this value is a five. But depending on the one after it, it could either stay the same or round up to a six. So let’s look at the next value. Well, this value is a nine. Nine is larger than five. Therefore, the value before it is going to round up and so this value is going to become a six. So to one decimal place, the mass of the displaced water is 4.6 kilograms.
