Video Transcript
A ship with a weight of 7000 newtons has both a motor and sails. The ship moves at a constant speed across the surface of the sea. The motor provides a force of 8500 newtons, and the drag from the water around the ship produces a force of 6000 newtons. The wind is blowing in the opposite direction to the motion of the ship, so the sails provide a drag force on the ship. What is the force on the ship from the wind pushing on its sails? How many newtons of force does the water underneath the ship apply upward on the ship’s hull?
Okay, so this is a question with two parts about the various forces acting on a ship. Let’s begin by drawing a diagram showing what all of these forces are. This here is our ship, and we’re told in the question that it has a weight of 7000 newtons. This weight force occurs as a result of gravity and acts vertically downward. Let’s label this force as 𝐹 subscript 𝑔, where the 𝑔 stands for gravity. At this point, we should recall that force is a vector quantity. And the important thing to remember about vector quantities is that, as well as having a magnitude, they also have a direction.
We’ve indicated the direction of the weight force using an arrow pointing downward in our diagram. Because forces have a direction as well as a magnitude, this means that we need to choose a direction to take as positive. When we’re talking about vertical forces, it’s usual to take the positive direction as being upward. With the upward direction taken as positive, the weight force of the ship, 𝐹 subscript 𝑔, is equal to negative 7000 newtons, where the negative sign is because this force acts vertically downward in the negative direction.
We should also pick a horizontal direction to take as positive. And for this, it makes sense to take the positive direction as being the direction of motion of the ship. Let’s imagine that the ship is moving in this direction to the right, so we’ll take right as the positive direction. In other words, forces acting to the right are positive and forces acting to the left are negative.
The question tells us that the motor provides a force of 8500 newtons. Since the motor is powering the ship’s motion, then the direction of the force provided by the motor must be in the direction of motion of the ship. So that’s to the right. We’ll label this force as 𝐹 subscript 𝑚. And we know that 𝐹 subscript 𝑚 is equal to 8500 newtons, where the value is positive because the force acts to the right.
We are also told that there’s a drag force from the water around the ship with a magnitude of 6000 newtons. This drag force is a result of the water resisting the motion of the ship. So the drag force must act to the left in the opposite direction to the ship’s motion. Let’s label this drag force from the water as 𝐹 subscript 𝑑. We have that 𝐹 subscript 𝑑 is equal to negative 6000 newtons, where it’s negative because it acts to the left.
These three forces are all of the forces that we’re told the values of. However, there are also two more forces that we need to add to our diagram. The question tells us that there’s a second drag force provided by the sails because there’s wind blowing in the opposite direction to the ship’s motion. We can deduce from this that the force provided by the sails must act in the opposite direction to the motion of the ship. So that’s to the left in our diagram. Let’s label this force as 𝐹 subscript 𝑠. And the value of 𝐹 subscript 𝑠 is exactly what we’re asked to find in the first part of the question.
The final force that we need to add to our diagram is the one referenced in the second part of the question. This is the upward reaction force applied by the water to the ship. So let’s add this upward reaction force to our diagram, and we’ll label it as 𝐹 subscript 𝑟.
So we’ve now got a diagram that shows all of the forces acting on the ship. The next step is to work out how we can use this information to find the values of the forces that we don’t currently know. To do this, we’re going to make use of Newton’s first law of motion. This law says that an object at rest remains at rest and an object moving with a constant velocity continues to travel at that velocity unless acted on by an unbalanced force.
What Newton’s first law of motion is telling us is that if no net force acts on an object, then that object will experience no change in velocity. We can also flip this logic around to say that if an object’s velocity isn’t changing, then there must be no net force acting on it.
The question text tells us that the ship moves at a constant speed across the sea. Since the ship is moving in a given direction with a constant speed, then it must have a constant velocity. As the ship’s velocity isn’t changing, then Newton’s first law of motion means that there must be no net force acting on it. What this means in practice is that the three horizontal forces acting on the ship must balance each other and the two vertical forces must also balance each other.
Let’s clear ourselves some space and see how we can use this information to answer the two parts of the question.
The first part of the question is about the force on the ship from the wind pushing on its sails. So this is asking us to find the value of the horizontal drag force we’ve labeled as 𝐹 subscript 𝑠. We’ve deduced from Newton’s first law of motion that there must be no net force acting on the ship and that this also means that there must be no net horizontal force on it. No net horizontal force means that the sum of all of the horizontal forces acting on the ship must equal zero.
The three horizontal forces are 𝐹 subscript 𝑚, the force provided by the motor, 𝐹 subscript 𝑑, the drag force from the water, and 𝐹 subscript 𝑠, the drag force from the wind pushing on the sails. This means that we have 𝐹 subscript 𝑚 plus 𝐹 subscript 𝑑 plus 𝐹 subscript 𝑠 is equal to zero newtons.
We’re trying to find the value of 𝐹 subscript 𝑠. So we need to rearrange the equation to make 𝐹 subscript 𝑠 the subject. To do this, we subtract 𝐹 subscript 𝑚 and 𝐹 subscript 𝑑 from both sides of the equation. On the left-hand side, the positive and negative 𝐹 subscript 𝑚 and 𝐹 subscript 𝑑 terms cancel each other out. This leaves just the 𝐹 Subscript 𝑠 term. On the right-hand side, subtracting these two terms from zero newtons simply gives us negative 𝐹 subscript 𝑚 minus 𝐹 subscript 𝑑.
Now that we have an equation where 𝐹 subscript 𝑠 is the subject, we’re ready to sub in our values for 𝐹 subscript 𝑚 and 𝐹 subscript 𝑑. When we do this, we’ll need to take some care over the negative signs. Subbing in the values, we get that 𝐹 subscript 𝑠 is equal to negative 8500 newtons minus negative 6000 newtons. These two negative signs end up negating each other to give us a plus. So we have that 𝐹 subscript 𝑠 is equal to negative 8500 newtons plus 6000 newtons, which works out as negative 2500 newtons.
Now it makes sense that this force is negative because in our diagram we’ve got it acting to the left, which we know is the negative direction. And this value of 𝐹 subscript 𝑠 is the force on the ship from the wind pushing on its sails, which is exactly what the first part of the question was asking us to find. So our answer to this first bit of the question is negative 2500 newtons.
Okay, let’s clear ourselves a bit of space to look at the second part of the question.
This second part is asking us for the upward reaction force on the ship. So that’s the value of the quantity we’ve labeled as 𝐹 subscript 𝑟 in our diagram. For the first part of the question, we used Newton’s first law of motion to deduce that there must be no net horizontal force acting on the ship. This same law also tells us that there must be no net vertical force acting on it. In this case, there are only two vertical forces acting. There’s 𝐹 subscript 𝑟, the reaction force of the water, and 𝐹 subscript 𝑔, the weight force.
Since we know that there is no net vertical force, then the sum of these two forces, 𝐹 subscript 𝑟 plus 𝐹 subscript 𝑔, must be equal to zero newtons. We’re trying to find the value of 𝐹 subscript 𝑟. So let’s make it the subject of the equation. We can do this by subtracting 𝐹 subscript 𝑔 from both sides. On the right-hand side, the positive and negative 𝐹 subscript 𝑔 terms cancel each other out. So we end up with an equation that says 𝐹 subscript 𝑟 is equal to negative 𝐹 subscript 𝑔.
Now that we’ve got this equation for the reaction force, 𝐹 subscript 𝑟, all we need to do is sub in our value for the weight force, 𝐹 subscript 𝑔. This gives us that 𝐹 subscript 𝑟 is equal to negative negative 7000 newtons. The two negative signs cancel each other out to give us a positive result of 7000 newtons, which makes sense because the reaction force is in the upward direction, which we took as positive. This upward reaction force applied by the water to the ship is exactly what we were asked to find. So our answer for this second part of the question is 7000 newtons.
