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Quote from: alancalverd on 31/01/2025 16:12:28If you tighten an M8 bolt to 10 Nm and continue to apply 10 Nm, will it turn another radian? No, it won't turn at all. If its rotation slows down, then the net torque is opposite of the direction of the rotation. If the rotational speed is constant, the net torque is zero. That includes the condition where it doesn't rotate at all. Your confusion comes from ignoring reactionary torque by the system.
If you tighten an M8 bolt to 10 Nm and continue to apply 10 Nm, will it turn another radian? No, it won't turn at all.
When someone says that the applied torque is 3.18 Nm, what they really mean is 3.18 Nm/rad.
Quote from: hamdani yusuf on 02/02/2025 14:31:35When someone says that the applied torque is 3.18 Nm, what they really mean is 3.18 Nm/rad.No they don't. The applied torque of a 3.18 N weight hanging from a 1 m lever is 3.18 Nm by definition. If you double the weight and nothing moves, what happened to the rad?
Here's a linear analogy for the situation.A 10 Newton weight is put on a table. It continuously apply 10 N force to the table. But it stays on the table instead of falling down, because there is normal force by the table to the weight, resulting in zero net force.
Quote from: hamdani yusuf on 01/02/2025 13:03:58Quote from: hamdani yusuf on 01/02/2025 10:15:34Here's a common values in industrial motor.P = 1000 W = 1000 Nm/sω = 3000 rpm = 50 rps ≈ 314 rad/sτ = P/ω ≈ 1000/314 ≈ 3.18 Nm/radYou can use an alternative unit, e.g.τ = P/ω = 1000/50 = 20 Nm/rotationThose values are physically the same magnitude of torque, only expressed in different units. Just like 0 degree Celcius equals 32 degrees Fahrenheit. Or 1 km/s equals 1000 m/s.When someone says that the applied torque is 3.18 Nm, what they really mean is 3.18 Nm/rad. It's NOT 3.18 Nm/rotation.It's NOT 3.18 Nm/degrees.
Quote from: hamdani yusuf on 01/02/2025 10:15:34Here's a common values in industrial motor.P = 1000 W = 1000 Nm/sω = 3000 rpm = 50 rps ≈ 314 rad/sτ = P/ω ≈ 1000/314 ≈ 3.18 Nm/radYou can use an alternative unit, e.g.τ = P/ω = 1000/50 = 20 Nm/rotationThose values are physically the same magnitude of torque, only expressed in different units. Just like 0 degree Celcius equals 32 degrees Fahrenheit. Or 1 km/s equals 1000 m/s.
Here's a common values in industrial motor.P = 1000 W = 1000 Nm/sω = 3000 rpm = 50 rps ≈ 314 rad/sτ = P/ω ≈ 1000/314 ≈ 3.18 Nm/radYou can use an alternative unit, e.g.τ = P/ω = 1000/50 = 20 Nm/rotation
Imagine a weightless tube, 2 m long, with a frictionless pivot in the middle and a 0.5 kg rocket nozzle at each end - an ideal firework, but powered by compressed air fed from the center. Turn on the air supply so each nozzle delivers 0.5N of thrust. The total torque is either 1 Nm or 1 Nm/rad.Describe the subsequent motion of the system using both definitions of torque.
No they don't. The applied torque of a 3.18 N weight hanging from a 1 m lever is 3.18 Nm by definition. If you double the weight and nothing moves, what happened to the rad?
Quote from: alancalverd on 02/02/2025 15:26:52Quote from: hamdani yusuf on 02/02/2025 14:31:35When someone says that the applied torque is 3.18 Nm, what they really mean is 3.18 Nm/rad.No they don't. The applied torque of a 3.18 N weight hanging from a 1 m lever is 3.18 Nm by definition. If you double the weight and nothing moves, what happened to the rad? What's the net torque? Is it non-zero? Your definition is not complete. It's only true if the direction of the force is perpendicular to the lever. Also, the position of the fulcrum or pivot point isn't changed by the force.
What would happen to the system if the torque is 1 Nm?
When I was making this double decker newton's cradle I released something very strange about levers.Oddly, the distance that seems to work well for long lever matches root 12 which matches the moment of inertia of a spinning rod *spinning from its end*. So it's as if we're matching the moment of inertia of just one side of the rod. That doesn't sit right with me though. Hmmm.
Then why am I leaning on the torque wrench, and why doesn't it fly backwards when I release it?
Static and kinetic friction are two types of friction that act on objects in contact. The key difference between them lies in the state of motion of the objects:Static Friction: * Acts on objects at rest: Static friction is the force that prevents two surfaces from sliding against each other when they are stationary relative to each other. * Variable force: It can vary in magnitude, increasing or decreasing depending on the force trying to initiate movement. * Has a maximum value: Static friction has a limit. Once the applied force exceeds this limit, the object will start to move.Kinetic Friction: * Acts on objects in motion: Kinetic friction is the force that opposes the motion of two surfaces sliding against each other. * Constant force: Unlike static friction, kinetic friction is generally constant regardless of the speed of the objects. * Lower than static friction: The force of kinetic friction is usually less than the maximum force of static friction. This means it takes more force to start an object moving than to keep it moving.Everyday Examples: * Static friction: Prevents a book from sliding off a tilted surface, keeps your shoes from slipping on the floor when you walk. * Kinetic friction: Slows down a sled sliding down a hill, allows you to apply the brakes on a bicycle.Mathematical Representation:Both static and kinetic friction forces are calculated using the following formulas: * Static Friction (Fs): Fs ≤ μs * N, where μs is the coefficient of static friction and N is the normal force between the surfaces. * Kinetic Friction (Fk): Fk = μk * N, where μk is the coefficient of kinetic friction and N is the normal force between the surfaces.Key takeaway:Static friction keeps things from moving initially, while kinetic friction opposes the motion of things already moving.
Stating the bloody obvious is not an explanation! That would require a statement of why sliding friction is less than static friction.
Friction coefficient, how to demonstrate the difference between static and dynamic friction in a physics experiment?
That's a great question! It seems counterintuitive that it's easier to keep something moving than it is to start it moving. Here's the breakdown of why sliding friction (kinetic friction) is less than static friction:1. The Nature of Surfaces: * Microscopic Interactions: Even surfaces that appear smooth to the naked eye have microscopic bumps and ridges. * Interlocking: When two objects are at rest, these irregularities have time to settle into each other, creating strong "interlocking" points. This interlocking is what static friction needs to overcome to initiate movement.2. Time Factor: * At Rest: When objects are at rest, the surfaces have time to conform and maximize their contact area, leading to more points of interlocking. * In Motion: When an object is already sliding, the irregularities don't have as much time to settle into each other. The "interlocking" is less pronounced. Imagine trying to quickly weave your fingers together versus doing it slowly and carefully ? the quick weave won't be as strong.3. "Breaking" vs. "Maintaining" * Static Friction: Static friction is about "breaking" those initial interlocking points. It takes more force to overcome this initial resistance. * Kinetic Friction: Kinetic friction is about "maintaining" motion. Once the object is moving, you're essentially just continuously breaking and reforming weaker, temporary bonds. This requires less force.In simpler terms:Think of it like pushing a heavy box. It takes a lot of effort to get it moving from a standstill (static friction). But once you have it sliding, it's easier to keep it going (kinetic friction) because you're not constantly having to overcome that initial "sticking" force.Key takeaway:The stronger interlocking of surfaces at rest leads to a higher static friction, while the constantly changing and less pronounced interlocking during motion results in a lower kinetic friction.
You really need to stop using AI.
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Quote from: Bored chemist on 04/02/2025 10:09:55You really need to stop using AI.why?
Quote from: hamdani yusuf on 04/02/2025 12:06:03Quote from: Bored chemist on 04/02/2025 10:09:55You really need to stop using AI.why?Two reasons.The first is that there's no guarantee that they don't talk nonsense.The second is neatly illustrated by the video you posted. The AI seems to have largely missed that factor.
The IssuePetition Summary:We call upon the International System of Units (SI) and global scientific bodies to standardize Joule per Radian (J/rad) as the official unit for torque, replacing the ambiguous Newton-meter (N?m). This change will eliminate confusion between torque and work/energy, improving clarity in physics, engineering, and education.Why This Matters:1. Ambiguity in SI Units ? Currently, both torque and work/energy are expressed in N?m, despite being fundamentally different quantities. Torque is a vector (rotational force), while work is a scalar (energy transfer). This leads to misinterpretations in academic and industrial contexts.2. Clarity in Scientific Communication ? Using Joule per Radian (J/rad) aligns torque directly with its definition: energy per unit angular displacement. This mirrors the linear counterpart (work = force ? displacement), making equations more intuitive.3. International Consistency ? Many fields, such as mechanical engineering and physics, already use J/rad in practical calculations. Formal adoption would unify scientific literature and teaching materials. Proposed Change:Officially define torque in Joules per Radian (J/rad) within the SI system.Encourage textbooks, research papers, and engineering standards to transition to the new unit.Join Us!Sign this petition to support a clearer, more logical unit system. Let?s urge ISO, NIST, BIPM, and other standardization bodies to take action!
Human potential will only accelerate with AI answering questions better and faster