# Tag Info

9

Then why none of such planet is torn apart till now ? Any planets and stars that would have or could have rotated themselves into pieces have presumably already done so (which practically means that they would never have been formed in the first place.) What we see around us today is from a human time scale a more or less stable state of the universe. ...

5

For example, if skater A pushes skater B, the lighter skater will accelerate further. Good enough so far. But you should be careful about the reasoning. Newton's third law says that when skater A pushes on skater B, there is an equal force applied by skater B to skater A (Mass isn't actually a consideration in Newton's third law). It's Newton's second law ...

3

I'm trying to figure out why they considered maximum static friction to act on the left block and then used that to find forces on the 4 kg block For $2kg$ : What is the magnitude of the external force acting on the $2kg$ block ? It is $10N$ and the maximum static friction that could act on it is just $8N$ , so we must consider this exact value of static ...

3

First, a hydrogen atom is not an exchange particle. It is a real particle. An exchange particle is termed a virtual particle. A virtual particle mediates an interaction between two other particles. The reason "virtual" is used is because it is usually "off mass shell" and not observable. The lifetime for such particles is ridiculously ...

3

The tensions are not the same because although the acceleration of all the masses is the same, the net force on each mass is different because the masses are different, making the tension different. Since all the masses are connected together their acceleration is the same, which you can obtain from Newton's second law using $F$ and the sum of the masses. ...

3

I don't understand how friction, for example, could have both a normal component and a frictional component. Your syllabus doesn't say this. It says that interaction forces have two components, normal and frictional. It isn't saying friction has these components; friction itself is one of the components. This is just a general application of breaking forces ...

2

Along with the tensile force, there will also be a pseudo force $2T\cos \theta$ in the downward direction. From the reference frame of the observer, the net force acting on $m_2$ mass is $2T\cos \theta$ in the downwards direction. Thus the acceleration of the block will be $2T\cos \theta \over m$ in the downwards direction. The pseudo force acting on $m_1$ ...

2

If I am understanding your questions correctly, just because the force of gravity and the centrifugal force are at equilibrium, it doesn't mean the shape has to be spherical. Think for instance of a water balloon flying through the vacuum that is also spinning. The internal force of the water trying to scape and the force of the rubber are at equilibrium but ...

2

So, the coefficient of kinetic friction is the magnitude ratio of the force of (kinetic) friction to the normal force. Your $mg\sin\theta$ theta term is not the force of friction, it's the magnitude of the parallel component of the gravitational force. It's not the case that the coefficient is dependent on the angle - an easy way to see this is to recognize ...

2

Impulse is defined as $F \Delta t$. So if you have a force $F$ which is applied to any object for a time $\Delta t$ you have an impulse regardless of the nature of the force. Whether it be gravitational or electrical or mechanical. Any force which causes a change in momentum and acts for a certain time can be termed an impulsive force. I think this term was ...

1

Thats simply the way compressive and tensile stress is defined. Stress isn't strictly a vector, having the same dimentions as pressure. So 'stress produced when the body is compressed' is called compressive stress, and 'stress produced when the body is stretched' is called tensile stress, even though as you said, the actual reaction forces are opposite.

1

Your diagrams are showing the directions of the external applied compressive and tensile stresses. In order to show the directions of the resulting internal compressive and tensile stresses you need to draw free body diagrams of cut sections of the beams. Those diagrams will slow the direction of the internal stresses as opposite to the direction external ...

1

The different parts of the universe we inhabit are causally-connected in the sense that one part (a cause) can interact with another part (the effect), by a variety of means at the atomic level. The details of those physical interactions don't matter for the moment. What does matter here is that none of those interactions are instantaneous; they are all ...

1

Your logic is solid. Indeed there is room for some ambiguity in the answer. The thing is tension and static friction both could be called self-adjusting forces. They adjust themselves according to other applied forces. In this case, both these forces are applied on a single object. There are 2 unknowns and 1 equation. $$F_0 = T +f_{static}$$ This system is ...

1

As you showed, the maximum frictional forces are enough to counter the external forces, so the center of mass of the system is definitely not moving. Since the string is inextensible, this means that the two blocks can't move in the opposite direction either. So the only natural thing to assume is that the blocks are at rest. For the first block to be at ...

1

The answer depends on the material of the body you are exerting force on. If it is a rigid body, then its shape is, by definition, constant. Assuming here for simplicity that your force vector passes through the COM, so that there isnt any torque on the body, then every particle will experience the same force in the direction you are pushing, as soon as you ...

1

A compass needle is magnetized and will experience a force or torque when placed in an external magnetic field such as one produced by a nearby current carrying wire. The compass needle is said to have a "magnetic moment" which measures how magnetized it is, and therefore how strongly it responds to external magnetic fields. Sometimes, magnetic ...

1

No , you are doing a mistake. If you take tension in each string to be equal to $F$ then for sure $M$ will not accelerate since $F$ gets balanced by the tension (say $T_1$) . Now $2M$ experiences $T_1$ and $T_2$ ($T_2$ from the other string) and according to you both have same magnitude $F$. So it should also not move. Now coming to the body of mass $4M$ , ...

1

Bare land is always going to hit you harder than water; if you've ever jumped into a swimming pool that was shallower than you thought it was, you'll know what I mean. But the video does say to aim for swamp, snow or trees, so that is going to narrow the difference, and I can imagine a landing in a tree might break your fall better than water. The force you ...

1

The reason why some things are better to fall on than others is because of how fast the momentum is reduced. Force can also be written as the 'rate of change of momentum', which means force is inversely proportional to the time period it is applied for. In other words, if your momentum is reduced from a very high number (something you'd expect if you're ...

1

The surface needs to support the weight of the body on it and hence exerts a normal. If the surface is a bit tilted, that is it makes an angle with the ground then it will exert a frictional force on a body lying on it to prevent the body from sliding down. The two components of the contact force are independent and that is the reason your textbook has ...

1

The difference is in the coordinate system. You are doing your calculation with $x=0$ as the point where the string is in its natural length. The equilibrium position, in that case, is $(m+m_c/2) g/k$. However, your professor is taking that equilibrium position $x_o=(m+m_c/2) g/k$ as his origin $x' = 0$ in the shifted coordinate system. You can arrive at ...

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