Video Transcript
A press is used to crack rocks, applying a pressure of 750 kilopascals to a rock with a mass of 6.2 kilograms, as shown in the diagram. The rock being crushed has a very uneven surface. The surface area in contact with the plates of the press is 20 centimeters squared. What is the average force applied to the rock by the press?
This first part of the question is asking us to work out the average force that the press applies to the rock. We’re told that this press applies a pressure to the rock of 750 kilopascals. We’ll label this pressure as 𝑝 so that we have 𝑝 is equal to 750 kilopascals. We’re also told that the surface area of the rock, which is in contact with the plates of the press, is equal to 20 centimeters squared. This surface area is the area over which the press applies a force to the rock. We’ll label it as 𝐴 so that we have 𝐴 is equal to 20 centimeters squared. We’ll label the average force applied to the rock as 𝐹. And this is the quantity that we’re trying to find.
We can recall that there is an equation which relates the three quantities pressure, area, and force. Specifically, pressure 𝑝 is equal to force 𝐹 divided by area 𝐴. In our case, we know the values of 𝑝 and 𝐴, and we want to find the value of the force 𝐹. So we need to rearrange this equation in order to make 𝐹 the subject. To do this, we take our equation and we multiply both sides by 𝐴. Then, on the right-hand side, the 𝐴 in the numerator cancels with the 𝐴 in the denominator. And we have that 𝐴 multiplied by 𝑝 is equal to 𝐹. We can also write this equation as 𝐹 is equal to 𝑝 times 𝐴, or in words force is equal to pressure multiplied by area.
Now, if we want to calculate a force in units of newtons, the SI base unit of force, then we need the pressure and the area in their own SI base units. So we need the pressure in units of pascals and the area in units of meters squared. Let’s start with the pressure. Currently, we have a value for pressure in units of kilopascals. This unit prefix of k means 1000, or in other words one kilopascal is equal to 1000 pascals. So then our pressure of 750 kilopascals is equal to 750 multiplied by 1000 pascals. This is equal to 750000 pascals. Alternatively, in scientific notation, we can write this pressure 𝑝 as 7.5 times 10 to the five pascals.
Now, let’s consider 𝐴, the surface area. We want to express this in units of meters squared. But currently we have a value in units of centimeters squared. We know that 100 centimeters is equal to one meter. If we divide both sides of this equation by 100, then we can see that one centimeter is equal to 0.01 meters. In scientific notation, we can write that one centimeter is equal to 10 to the minus two meters. If we then take the square of both sides of this equation, we have that one squared centimeter squared is equal to 10 to the minus two squared meters squared. One squared gives us one, and the square of 10 to the minus two gives us 10 to the minus four. So we have that one centimeter squared is equal to 10 to the minus four meters squared.
Going back to our area, 𝐴, of 20 centimeters squared, we see that this 20 centimeters squared is equal to 20 times 10 to the negative four meters squared. This gives us a value for 𝐴 in meters squared of two times 10 to the minus three meters squared. So we now have a value for the pressure 𝑝 in units of pascals and a value for the area 𝐴 in units of meters squared. This means that we’re ready to substitute these values into this equation to calculate the force 𝐹 and that the force we calculate will have units of newtons.
Substituting these values gives that 𝐹 is equal to 7.5 times 10 to the five pascals multiplied by two times 10 to the minus three meters squared. Then, 7.5 multiplied by two gives us 15. And 10 to the five multiplied by 10 to the negative three gives us 10 to the two. And we already said that if the pressure was in pascals and the area was in meters squared, then the force will be in units of newtons. We can also write this force of 15 times 10 to the two newtons as 1500 newtons. And so our answer to this first part of the question is that the average force applied to the rock by the press is equal to 1500 newtons.
Now, let’s look at the second part of the question.
How many times heavier would the rock being crushed need to be for it to apply the same weight to the ground as the average force applied by the press? Round your answer to one decimal place.
Okay, so we’ve already calculated the average force applied by the press in our answer to the first part of the question. We know that this is equal to 1500 newtons. We now need to find how many times heavier the rock would need to be in order to apply the same weight to the ground as this average force. We can recall that weight, which we’ve labeled 𝑊, is equal to mass, which we’ve labeled as 𝑚, multiplied by 𝑔, where 𝑔 is the acceleration due to gravity and has a value of 9.8 meters per second squared. We are asked to consider the situation where this weight is equal to the average force applied to the rock by the press. In other words, the weight 𝑊 is equal to 1500 newtons.
Since we know that 𝑊 is equal to 𝑚 multiplied by 𝑔, then in this equation we can replace 𝑊 by 𝑚 times 𝑔 to say that 𝑚 times 𝑔 is equal to 1500 newtons. If we divide both sides of this equation by 𝑔, then on the left-hand side, the 𝑔 in the numerator cancels with the 𝑔 in the denominator. Then, we can substitute in that 𝑔 is equal to 9.8 meters per second squared. Evaluating the right-hand side gives us that 𝑚 is equal to 153.06 kilograms, where the ellipses indicate that there are further decimal places. We get a mass with units of kilograms, the SI base unit for mass, since all the other quantities in our calculation were in their SI base units.
Physically, this value of 𝑚 is the mass of a rock which would apply a weight of 1500 newtons to the ground. Now, we weren’t actually asked to work out this value of mass, but rather how many times heavier the rock being crushed would need to be in order to apply this weight of 1500 newtons. We’ve just worked out that to apply this weight, the rock would need a mass of 153.06 kilograms. Meanwhile, we’re told in the question that the rock in the press actually has a mass of 6.2 kilograms.
We’ll label this mass as 𝑚 subscript r, where the r stands for rock. What we’re trying to work out is how many times bigger this value of 𝑚 subscript r would need to be in order to be equal to this required mass of 𝑚. We could formulate this statement mathematically as 𝑚 subscript r multiplied by 𝑥 is equal to 𝑚, where 𝑥 is the quantity that we’re trying to find. In other words, 𝑥 is how many times bigger 𝑚 subscript r would need to be in order to equal 𝑚. If we then divide both sides of this equation by 𝑚 subscript r, then on the left-hand side, the 𝑚 subscript r in the numerator cancels with the one in the denominator. And we’re left with an equation that says 𝑥 is equal to 𝑚 divided by 𝑚 subscript r.
We can then substitute in the values of 𝑚 and 𝑚 subscript r. Evaluating this right-hand side gives us that 𝑥 is equal to 24.687, where the ellipses indicate that there are further decimal places. If we look back at the question, we see that we are asked to round our answer to one decimal place. To one decimal place, our result rounds up to 24.7. So our answer to the second part of the question is that the rock would need to be 24.7 times heavier in order to apply the same weight to the ground as the average force applied to the rock by the press.
