Dynamometer Reading: Submerged Object In Water Calculation

Introduction

Hey guys! Today, we're diving deep into a classic physics problem involving buoyancy, gravity, and dynamometers. This is the kind of stuff that might seem tricky at first, but once you break it down, it's actually pretty cool. We're going to figure out what a dynamometer will read when an object is submerged in water. This involves understanding the forces at play, including the weight of the object, the buoyant force acting upwards, and how these forces ultimately affect the tension in the spring of the dynamometer. So, let's get started and make sure we understand each concept thoroughly. We’ll go step-by-step, so you can follow along easily. Remember, physics is all about understanding the why behind the what, so let's get ready to learn something awesome today!

Problem Statement

Okay, so here’s the scenario. We've got an object with a mass ( ${m}$ ) of 0.7 kg and a volume ( ${V}$ ) of 200 cm³. This object is hanging from a dynamometer – think of it like a fancy scale that measures force – and it’s completely submerged in water. The water has a density ( ${\rho}$ ) of 1 g/cm³. Our mission, should we choose to accept it, is to figure out what the dynamometer is going to read. What force will it display? This is where physics gets interesting because we need to consider all the forces acting on the object, not just its weight. We've got gravity pulling the object down, but we also have the buoyant force of the water pushing it up. The dynamometer reading will reflect the net force – the result of this tug-of-war. So, let’s put on our thinking caps and dive into the solution!

Understanding the Forces Involved

Before we jump into calculations, let's break down the forces at play here. This is super important, guys, because if you don't understand the forces, you'll be swimming in circles (pun intended!).

1. Weight (${W}$): This is the force of gravity pulling the object downwards. We calculate it using the formula:

   ${W = mg}$

   where ( ${m}$ ) is the mass of the object and ( ${g}$ ) is the acceleration due to gravity (approximately 9.8 m/s²).

2. Buoyant Force (${F_B}$): This is the upward force exerted by the water on the object. It's equal to the weight of the water displaced by the object (Archimedes' principle). We calculate it using the formula:

   ${F_B = \rho V g}$

   where ( ${\rho}$ ) is the density of the water, ( ${V}$ ) is the volume of the object (which is equal to the volume of water displaced), and ( ${g}$ ) is the acceleration due to gravity.

3. Tension (${T}$): This is the force exerted by the dynamometer's spring, and it's what the dynamometer reading will show. The tension force acts upwards, opposing the net downward force (which is the weight minus the buoyant force).

So, to recap, we’ve got weight pulling down, buoyant force pushing up, and tension trying to keep everything in equilibrium. The dynamometer measures this tension. Make sense? Great! Let's move on to the calculations.

Step-by-Step Calculation

Alright, guys, let's get our hands dirty with some numbers! This is where we'll use the formulas we just talked about to actually calculate the dynamometer reading. Don't worry, we'll take it one step at a time.

1. Calculate the Weight (${W}$)

We know the mass ( ${m}$ ) of the object is 0.7 kg and the acceleration due to gravity ( ${g}$ ) is approximately 9.8 m/s². Let’s plug these values into our weight formula:

${W = mg = 0.7 \text{ kg} \times 9.8 \text{ m/s}^2}$

${W = 6.86 \text{ N}}$

So, the weight of the object is 6.86 Newtons. That’s the force pulling the object downwards.

2. Calculate the Buoyant Force (${F_B}$)

Here, we need the density of water ( ${\rho}$ ), the volume of the object ( ${V}$ ), and, of course, gravity ( ${g}$ ). Remember, the density of water is given as 1 g/cm³, but we need to convert it to kg/m³ to keep our units consistent. Since 1 g/cm³ is equal to 1000 kg/m³, we’ll use that value.

Also, we need to convert the volume from cm³ to m³. Since 1 m³ = 1,000,000 cm³, we have:

${V = 200 \text{ cm}^3 = 200 \times 10^{-6} \text{ m}^3 = 2 \times 10^{-4} \text{ m}^3}$

Now, we can calculate the buoyant force:

${F_B = \rho V g = 1000 \text{ kg/m}^3 \times 2 \times 10^{-4} \text{ m}^3 \times 9.8 \text{ m/s}^2}$

${F_B = 1.96 \text{ N}}$

So, the buoyant force pushing the object upwards is 1.96 Newtons.

3. Calculate the Tension (${T}$)

The tension in the dynamometer string is what we’re trying to find. The tension force balances the net downward force, which is the weight minus the buoyant force:

${T = W - F_B}$

${T = 6.86 \text{ N} - 1.96 \text{ N}}$

${T = 4.9 \text{ N}}$

Therefore, the dynamometer will read 4.9 Newtons. Woohoo! We did it!

Final Answer and Interpretation

Okay, guys, so we've crunched the numbers, and here’s what we found: the dynamometer reading is 4.9 Newtons. But what does this actually mean? Well, it tells us the amount of force the dynamometer needs to exert to hold the object in equilibrium while it’s submerged in water. Think about it: the object's weight is trying to pull it down, but the water is pushing it up with the buoyant force. The dynamometer is just there to make up the difference and keep everything stable.

If we had simply hung the object in the air, the dynamometer would have read 6.86 N (the object's weight). But because we submerged it in water, the buoyant force helped to support some of the weight, reducing the load on the dynamometer. This is a classic example of how buoyancy affects the forces we measure in the real world. Pretty neat, huh?

Common Mistakes to Avoid

Alright, let's talk about some common pitfalls that students often stumble into when solving problems like this. We want to make sure you guys are smooth sailing, so let's steer clear of these errors!

1. Forgetting to Convert Units: This is a big one! We dealt with grams, cubic centimeters, and kilograms all in one problem. If you don't convert everything to a consistent set of units (like meters, kilograms, and seconds), your answer will be way off. Always double-check your units before plugging them into formulas.
2. Mixing Up Density Values: Make sure you're using the correct density! In our case, we used the density of water. If the object were submerged in a different liquid, you’d need to use that liquid’s density instead.
3. Ignoring Buoyant Force: This is the most common mistake. People sometimes forget that when an object is submerged, the water is exerting an upward force. Always consider the buoyant force in these types of problems.
4. Incorrectly Calculating Buoyant Force: Remember, the buoyant force depends on the volume of the object (or the volume of water displaced), not the object's density. Double-check that you're using the correct volume in your calculation.
5. Misunderstanding the Question: Sometimes, the question might ask for the buoyant force itself, or the apparent weight (which is what the dynamometer reads). Make sure you're answering the specific question being asked.

By keeping these common mistakes in mind, you'll be much better equipped to tackle similar problems with confidence!

Real-World Applications

Okay, guys, so we've solved a cool physics problem, but you might be wondering, “Where would I ever use this stuff in real life?” Well, the principles we've discussed today – buoyancy, force measurement, and equilibrium – are actually used in tons of different applications. Let's explore a few examples:

1. Naval Architecture: When engineers design ships and submarines, they need to be experts in buoyancy. They need to calculate the buoyant force to ensure the vessel floats stably and can carry the intended load. Understanding these principles is crucial for safe and efficient maritime transportation.
2. Hot Air Balloons: Hot air balloons float because the hot air inside the balloon is less dense than the cooler air outside. This creates a buoyant force that lifts the balloon. The principles we’ve discussed help explain why and how hot air balloons work.
3. Underwater Construction: When constructing underwater structures like bridges or pipelines, engineers need to account for the buoyant force on the materials and equipment. This is vital for planning the construction process and ensuring the stability of the final structure.
4. Measuring Density: The concept of buoyancy is also used in devices that measure the density of liquids. By observing how an object floats in a liquid, we can determine the liquid’s density – a useful tool in many scientific and industrial applications.
5. Diving and Scuba Gear: Scuba divers use buoyancy control devices to adjust their buoyancy underwater. Understanding buoyancy helps divers maintain their depth and move efficiently through the water.

So, as you can see, the physics we've discussed today isn't just theoretical stuff. It has real-world implications that impact various industries and technologies. Keep an eye out, and you'll start noticing these principles in action all around you!

Practice Problems

Alright, guys, now that we've gone through the theory and calculations, it's time to put your knowledge to the test! Practice makes perfect, so let's try a few more problems to solidify your understanding. Here are a couple of scenarios for you to tackle:

1. Problem 1:

   A stone with a mass of 1.5 kg and a volume of 500 cm³ is submerged in a liquid with a density of 1.2 g/cm³. What will a dynamometer read if the stone is suspended from it and fully immersed in the liquid?

2. Problem 2:

   An object weighs 10 N in air. When fully submerged in water, it appears to weigh only 6 N. What is the volume of the object?

Take your time to work through these problems. Remember to follow the steps we discussed: identify the forces, calculate the weight and buoyant force, and then determine the tension or the desired variable. Don't be afraid to go back and review the concepts if you get stuck. Physics is all about building your understanding step by step.

After you've attempted these problems, try to explain your solutions to a friend or family member. Teaching someone else is a fantastic way to reinforce your own knowledge. Good luck, and happy problem-solving!

Conclusion

So, guys, we've reached the end of our deep dive into buoyancy, dynamometers, and submerged objects! We’ve covered a lot today, from understanding the forces at play to calculating the dynamometer reading and exploring real-world applications. Remember, the key to mastering physics is to break down complex problems into smaller, manageable steps. We looked at how to calculate the weight of an object, the buoyant force acting on it when submerged in water, and the resulting tension measured by a dynamometer. We also discussed common mistakes to avoid and saw how these principles are used in various fields, from naval architecture to diving.

I hope this explanation has been helpful and has sparked your curiosity about the world of physics. Keep practicing, keep asking questions, and most importantly, keep exploring! Physics is all around us, and with a little bit of effort, you can unlock its secrets. Until next time, keep those brains buzzing and stay curious! You've got this!