Question Video: Determining Force on a Submerged Surface | Nagwa Question Video: Determining Force on a Submerged Surface | Nagwa

# Question Video: Determining Force on a Submerged Surface Physics • Second Year of Secondary School

Water in a container is pushed horizontally by a movable wall of the container, as shown in the diagram. Inside the water-filled part of the container is a square metal sheet with size of length 0.25 m. The base of the sheet is attached to the floor of the container, keeping the sheet vertically oriented. The water pushes the sheet when the container’s movable wall is moved. The movable wall has sides of lengths 𝐿₂ = 0.25 m and 𝐿₃ = 0.75 m. The force applied to the movable wall is equal to the force applied to the water by the wall. What is the magnitude of the force that pushes the surface of the metal sheet that faces the movable wall?

04:12

### Video Transcript

Water in a container is pushed horizontally by a movable wall of the container, as shown in the diagram. Inside the water-filled part of the container is a square metal sheet with size of length 0.25 meters. The base of the sheet is attached to the floor of the container, keeping the sheet vertically oriented. The water pushes the sheet when the container’s movable wall is moved. The movable wall has sides of lengths 𝐿 two equals 0.25 meters and 𝐿 three equals 0.75 meters. The force applied to the movable wall is equal to the force applied to the water by the wall. What is the magnitude of the force that pushes the surface of the metal sheet that faces the movable wall?

Looking at our diagram, we see this square metal sheet vertically oriented in our container with water in it. One of the four sides of this container is a movable wall, and we see that a force of 75 newtons is being applied to that wall. 𝐿 two and 𝐿 three are the dimensions of that movable wall that are submerged in water. In this part of our question, we want to figure out the magnitude of the force that pushes on the surface of the metal sheet that faces the movable wall. The direction of that force then will be to the right away from that wall.

To begin on our solution, let’s note that this force applied over this area of our movable wall creates a pressure. By Pascal’s principle, that pressure is equal to the force divided by the area. The pressure 𝑃 created by this force of 75 newtons is then transmitted all throughout the fluid in the container. That pressure then acts on any object in the container, such as the square metal sheet. Since the force we’re interested in solving for acts to the right, let’s call this force 𝐹 sub 𝑅. The pressure 𝑃 created by our force acting on the movable wall equals that force, we’ll call it 𝐹 sub wall, divided by the area of the wall that is underwater.

We mentioned that this pressure is transmitted all throughout the water in the container. Therefore, 𝑃 is also equal to the force we want to solve for, 𝐹 sub 𝑅, divided by the area of the square metal sheet. Rearranging this equation, that force is equal to the ratio of the area of the sheet to the area of the wall that’s submerged multiplied by the force that acts on the wall. To help us solve for this result, let’s clear some space at the top of our screen. And let’s note that 𝐹 sub wall, it’s called simply 𝐹 in our diagram, is 75 newtons, 𝐿 sub two is given in our problem statement as 0.25 meters, 𝐿 sub three is 0.75 meters, and lastly each of the sides of our square metal sheet are 0.25 meters long. This means the area of the sheet is this dimension squared.

We’re nearly ready to calculate the force 𝐹 sub 𝑅. Before we do, let’s note that the area of the wall is equal to 𝐿 two times 𝐿 three. Therefore, this is how we’ll calculate 𝐹 sub 𝑅. Note that in this equation the units of meters squared in the numerator cancel with meters squared in the denominator so that we’ll end up with the units simply of newtons, units of force. Calculating this expression, we find a result of 25 newtons. This is the magnitude of the force acting on the square metal sheet on the side that faces the movable wall.

Let’s look now at part two of our question.

What is the magnitude of the force that pushes the surface of the metal sheet that faces away from the movable wall?

So if this is a side-on view of our metal sheet, in part one of our question, we calculated the force acting to the right on this sheet, 𝐹 sub 𝑅. In this second part, we want to calculate the force that acts on the surface of the metal sheet that faces away from the movable wall. That force would push on the metal sheet to the left, so we’ll call it 𝐹 sub 𝐿. To find 𝐹 sub 𝐿, we actually won’t need to do any calculations.

Recall from earlier we’ve said that the pressure created by this movable wall pushing on the fluid is transmitted equally all throughout that fluid. This means that any point in the fluid, say, this point over here, will experience equal forces acting in all directions. The same holds true for any point on the surface of the metal sheet and indeed the forces on the metal sheet themselves. These forces all balance out equally acting in all directions. If this were not the case, then the metal sheet would experience a net force. It would accelerate. But as the metal sheet is in equilibrium, we can say that the magnitudes of these two forces are equal. Therefore, just as 𝐹 sub 𝑅 was 25 newtons, so is 𝐹 sub 𝐿.

The magnitude of the force that pushes on the surface of the metal sheet that faces away from the movable wall is 25 newtons.

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