# Tag Info

1

Say you have a weight tied to each side a a rope which is strung over a pulley with friction. Here's a really easy way to see why the tensions on each side of the rope can't be equal. Imagine a really stiff pulley - in other words, ${\bf F}_{friction}$ is high. If that's the case, it'll be possible to balance unequal loads on this pulley system - i.e. a ...

1

We would have to know more about the ice we want to build the wall with. For example, for ice in icesheets, you have an ice which effectively reaches a plastic region of the stress strain curve at around $0,5 MPa$. I am not a geologist, but I believe that the glaciers can be only thicker than $\sim 50 m$ thanks to it's specific shape and the fact that the ...

1

It might not be best to say you'd feel it, as the oscillations of the sun's gravity (as due to rotations and revolutions) are relatively subtle here on earth. The moon's gravity has twice as much influence in terms of tidal force, which is probably the most prominent consequence of gravitational "stacking" in the macroscopic sense you had in mind.* Here's an ...

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Well given Newtons famous equation the force of gravity you feel is equal to $$F_{\textrm{gravity}}=G\frac{mm'}{r^2}.$$ It increases with the mass of the two objects being measured and is inversely proportional to the distance between the two objects in question. Strangely enough however, this technically stretches out nearly infinitely so the ...

1

Technically, yes, however in the case of you standing on a planet: You feel the gravity of the planet and star, so you accelerate accordingly. However the planet also feels the gravity of the star, so accelerates towards it. So you only notice your acceleration towards the planet.

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You can analyze this by imagining a tiny tiny gap between the two masses. Physically we have exactly that, as the electrons at the surface of the first block are certainly not in contact with the electrons at the surface of the second block. Then we have a series of two collisions: the first is the initial impulse, the second is the collision between the two ...

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I'm going to just use $32^∘$ below; it doesn't make a difference. Your equation isn't correct. You should have $F_x−f_k = 0$ or $Fcos32∘−f_k = 0$. The x-component of F is $F_x = F.cos32^∘$. Writing $F_x.cos32^∘$ doesn't make logical sense. Why? Shouldn't I be able to use $∑F=0$ in this problem to find the answer? Yes, you can.

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You must obtain the magnetic field due to the solenoid according to the approximate calculations according to its dimensions. If the magnet is small you can easily found the torque, and also the force when it is in a region of non uniformity.

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You have a chain of action and reaction. There is twice the weight on our head because the forces felt by the lower block are its weight, plus the action of the top block (minus the reaction force of your head). Then how is it on a molecular scale? Well, just the same: if you imagine a crystalline solid with horizontal layers, one layer feels the action of ...

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You're mixing up different things. Two blocks of iron press your head more than one block because the Earth's gravity pulls two blocks stronger than one. This is why two blocks can tear through the paper where one can not. The molecular bit comes into consideration if you ask why the top block doesn't pass through the bottom one. This is because the ...

2

You should read the article in wikipedia on nuclear force. Various models exist that describe the behavior of nuclear forces, which are the result of a spill over of the strong force, the force that exists within the proton and the neutron. From the link Force (in units of 10,000 N) between two nucleons that experience the nuclear force, as a ...

0

If your friend's energy+ yours= F1, then you would see your own energy expenditure halved, which we know cannot be the case. If your friend helps you push the object, then you are no longer applying the same force, or (lazy answer) the force is no longer localised and motivates the part of the object most subject to friction. So, once the object is first ...

2

There is, effectively, only gravitation and friction acting on the pack of gum. However, the friction is not that strong (it is mostly independent of the velocity of the book, and dynamic friction is weaker than static friction) and it doesn't have that much time to act. Hence it doesn't affect the momentum of the gum noticeably. This is very related to the ...

3

I'm sure everyone has had that concern when we encountered the definition for the first time, in school. There is a valid reason why this definition is still persisted with, despite the deficiency that you hit on. The most popular (and simple) forces in physics (also the ones with which we begin learning physics) are conservative forces, implying that the ...

32

If you're pushing a 10-ton truck and it's not moving, you are not doing any work on the truck because the distance $ds=0$ and the nonzero force $F$ isn't enough for the product $F\cdot ds$ to be nonzero. Your muscles may get tired so you feel that you're "doing something" and "spending energy" but it's not the work done on truck. You're just burning the ...

1

Equations of motion explained: Sum of forces on body equals mass times acceleration of center of mass. $$\sum \vec{F} = \frac{{\rm d} }{{\rm d}t}(m\vec{v}_{cm}) = m \vec{a}_{cm}$$ Sum of moments on center of mass equals mass moment of inertia times angular acceleration of body, plus velocity related terms. $$\sum \vec{M}_{cm} = \frac{{\rm d} }{{\rm ... 3 Many students confuse the term work in physics with the conventional term of work. Your body wastes energy when you push something, and when that something doesn't move... 100% is wasted in the biological efficiency. 1st step: forget the concept of how hard it would be for you to do it. How much work is a table doing by holding up a 1kg weight? zero. It ... 3 The same force applied anywhere on a rigid object causes the same translational accelleration. The difference is that forces not applied in the direction of the center of mass will also cause some rotational accelleration. Remember that F and A in F=mA are vectors. You can therefore treat the vectors as components in any orthagonal system you like. One ... 0 It is easy if you draw the Free body diagram of the box. The FBD should show only the EXTERNAL forces. There are 2 external forces. First, the force of the earth's gravity ON the box which equals Mg and acts in a downward direction (here g = +ve). Second the force of the table ON the box which is N acting upwards. Now you decide on your sign ... 3 In a sense your question is entirely apropos. The critical distinction to be made, however, is that the two forces can only be summed to zero if we are talking about the two-body system. Each body, considered in isolation, experiences an unbalanced force and thus experiences accelerated motion. The system, however, only experiences mutually-canceling ... 0 This may be cheating, but I think the problem is easier if you use conservation of energy. If you set the gravitational potential energy reference to the height of m1, then initially you have$$ U_i = k\frac{q^2}{d}. $$The final energy will have a gravitational potential energy and an electrical potential energy:$$ U_f = mgh + k\frac{q^2}{r} $$A bit of ... 0 I worked the problem out a ways and it involves quite a bit of tedious algebra. The technique I used was to include the electric force in the Fx and Fy equations by looking at the angle that  \hat{r} (the vector between the two charges) makes between the charges. For example, the equation i came up for Fx is T_x - \frac{Kq^2}{r^2} \cos{\theta} =0 where ... 0 This is the classic horse and buggy brain teaser. If the horse pulls on the buggy, the buggy must pull back on the horse with an equal force so who moves anywhere? The answer lies in the fact that the buggy pulls back on the horse because it is being ACCELERATED. That is where the force comes from. In the case of the two boxes, the second box (one at the ... 1 A better electrical analogy to Newton's second law might be inductance:$$ V = L \frac{\mathrm d i}{\mathrm d t} $$The only reason this looks different is that physics has a name and conventional symbol for the derivative of speed, but electronics does not have a name for the derivative of current. So let's just pretend that the word acceleration does not ... 0 I'm not sure if I'm just misinterpreting your question, but we can most certainly influence the acceleration of an object directly. To do this we must take into consideration what is actually happening as we accelerate; Acceleration, in physics, is the rate at which the velocity of an object changes over time. While it is a sum of the net forces over the ... 1 We should think a bit more carefully what force, mass and acceleration really are: For simplicity, we consider a classical point particle. The force \vec F is something externally applied to the particle, it is a property of its environment. Most often, it is the gradient of a potential, \vec F = - \nabla V, but it need not be. In general, it is some ... 4 Your observations are spot on. I usually write Newton's second law this way: \vec{a} = \vec{F}/m. This form makes it clear that the law is a relationship between the dynamic variables force and mass, and the kinematic variable, acceleration. F and m describe the situation, a is the result. Cause and effect, if you will. In fact, that's one ... 1 In my view, an expression like V=RI, practically speaking, isn't much different. You have to be concrete with a real-world example. I can connect a resistor across a voltage source, and then I can only alter the voltage or change the resistance value to change the current; I cannot change the current directly in this configuration. Your statement that ... 0 If the acceleration (presumably measured on the sled) is the same, then the mass of the sled has no effect. It will take 40 times a s much force to accelerate the 40 lb sled. What is more likely to be happening is that at a particularly frequency, then mass of the cap and the spring constant of the legs will be such that you get metal fatigue from the ... 2 When you take a brass plate of considerable thickness and place it in between two charges, say positive and negative, induction takes place in the brass plate since it is a conductor: the electrons shift to the end near the positive charge while the cations stay near the negative charge. Now, induction occurs in order to make the field outside a certain ... 1 The brass plate is a conductor, so the potential will be the same on both sides. The thickness of the brass plate therefore subtracts from the effective distance between the two charges, making the electric field strength higher in the remaining open space between the charges. This stronger field will cause more force to be experienced by each charge. ... 0 A known mass tied to a thread will exert a constant force, proportional to its mass. That's the easiest of all. A spring under a light load will create a constant force that is proportional to the extension. See Hooke's law. A magnet and a piece of magnetic iron (or other magnet) tied to a known distance will do, too. If it is an electro-magnet, the force ... 1 There is no quick answer, except if the droplet is completely non-wetting or if it is at least partly wetting. If it is completely non-wetting, it will be move towards the wide side of the funnel until it is a spherical drop touching only its wall. If it is at least partly wetting, it will move to the narrow side until it reaches its apex (if air is ... 0 in this situation a normal man will have less weight than horse and car so it clearly states that acceleration must be same. No. If the force is the same for all 3, then if the mass is different, the acceleration must be different. Since the man has the least mass, then he will accelerate the fastest. So the man would win the race. But in the real ... 2 Let's work through this problem...$$\text{Power} = \frac{\text{Work}}{\text{Time}} \; \text{and} \; \text{Work} = \text{Force}\cdot \text{Displacement} $$therefore$$\text{Power} = \frac{\text{Force}\cdot\text{Displacement}}{\text{Time}} $$From Newton's Second law we know$$\text{Force} = \text{Mass}\cdot\text{Acceleration} Substituting again: ...

0

Real bodies aren't airtight. In fact, one of the most important principles guiding the evolution of multicellular organisms is that materials (such as oxygen) need to be exchanged throughout the organism, and so many interfaces are actually conducive to equilibrating pressure. When the ambient air pressure increases, the air pressure in your lungs increases ...

1

By way of analogy, think of what happens when you blow up a balloon and let it go. It spins around, goes this way and that. A balloon rarely goes straight, without spinning. The thrust from a balloon rarely goes through the center of mass. It rotates and translates. Because the thrust vector itself turns with the rotating balloon, the translation is not ...

0

Apparently it's better to ask questions only one-at-a-time on this site, but I understand why you did it like that; these are all related. Now, these are some really interesting questions, hopefully I can help you out (there's some interesting subtleties to consider due to the thrusters). Answer 1 If you fire only B, then it will rotate, and it will ...

0

It is due to Archimedes' principle. The total force on the 'air bubble' under the water is the force of gravity on it (downwards) plus the force of gravity on the water that the 'air bubble' displaces. Since the same volume of water that replaces the air is more massive (because of higher density) there is a net upward force on the bubble and causes it to ...

2

Perhaps an easy way to see where Newton's second law comes from is as follows. Imagine an object sitting in space (so no friction, etc. to worry about). You can push on it (exert some force on it), and see what happens to it. When you push on it with a constant force, you see the object start to accelerate, which means its velocity increases linearly with ...

0

You are mostly correct. Since you are only asked for a qualitative result, you can simplify what you said: If the boy pulls on rope with force $F$ for time $t$, it will move a certain distance $d$. If the other end of the rope is fixed, the work done is $F\cdot d$. If the end of the rope is moving towards him, the total length of rope he reels in is greater ...

0

The magnetic field created by the wire is azimutal while the field created by the solenoide can be studied as the sum of two contributions. The first one is an uniform longitudinal field inside the solenoid and the second is an azimutal field outside the solenoid that varies as I/r. Therefore the magnetic force the wire experiences is null because current ...

1

Consider this container in pressurized air but zero-gravity (and ignore surface tension, which would make the liquid ball up). If your guess were right, the liquid would squirt out the small hole on the right, but that ignores the role of the wall on the right, which counters the pressure on the left. Think of the liquid as a collection of horizontal ...

0

In short, you are right, the pendulum has an angle with respect to the gravity vector, but your reference frame has this same angle, making then aligned. A 1-D coordinated turn is a turn such that the horizontal component of the lift is equal and opposite to the force generated by centrifugal acceleration. E.g like this: Obtained from Free online pilot ...

5

For the case that you have drawn, the behavior of the drop is actually the exact opposite of what you mention: it will move from right to left. This is caused by surface tension and the curvature of the droplet caps which creates a larger pressure in the drop at side B than at side A. To make it more quantitative. Let's assume that the funnel is ...

2

I think I see your question, if the surface of the air bubble were perfectly flat, and the air + cup didn't float, then the surface would have an equal pressure across it and it would not move. The system, however, is in an unstable equilibrium, the slightest perturbation will cause the bubble to rise out of the cup. Consider this, since the pressure of the ...

1

When the cup is tilted up, the water wants to flow into the cup. That is what water does - it attempts to flow downstream. In doing so it displaces the air. Now the air experiences the force of the water (pressure below bubble > pressure above)

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Like you said, the water is pushing UP. It is always pushing the air up (it is the definition of buoyancy). Therefore, when you flip the cup, the buoyancy force continues to push the air upwards. Before you flipped the cup, the cup is exerting the downward force that counteracts the buoyancy force. Therefore in the FBD, the buoyancy force (pointed upwards) ...

0

I'm quite confused about your calculation. E.g. I think $a=\theta$. However, to solve this problem, the proper Langrangian is: $L=T-V=\frac12 m_1 \dot{r}^2+\frac12m_2(\dot{r}^2+r^2\dot{\theta}^2)-gr(m_1-m_2\cos(\theta)))$ You can find the e.o.m. by solving the Euler-Lagrange eqns.: \$0=\frac{\partial}{\partial t}\frac{\partial L}{\partial ...

1

Equations for magnetic interactions between objects tend to be a lot more complicated than for electrostatic forces. This is because while electric fields are produced by and exert forces on charge, a scalar, magnetic fields interact with electric currents, the flow of charge in a particular direction, which is a vector quantity. This makes the equations for ...

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