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14

Not to detract from Floris' answer, but I think this is an instance where it is nice to think in terms of limits. If the hay is tied down, you're stopping an object with mass (truck + hay). If the hay isn't tied down, but on a sufficiently sticky surface such that it doesn't move, it should be the same as stopping it if it were fixed, since the outcome is ...


13

On the whole, static friction is higher than dynamic friction. This means that if you can brake without your wheels skidding, you will come to a halt more quickly. So let's assume that the truck brakes without skidding, and see where that gets us. Let's assume that your truck has weight $W = Mg$ with a haystack with additional weight $w = mg$ on top. ...


6

In general relativity, rather than a two objects exerting a gravitational force on each other, the two objects are both part of the stress-energy tensor. This tensor determines the shape of spacetime (via the spacetime metric), and the spacetime metric determines what the geodesics are (roughly speaking, the metric determines how an object will move when no ...


5

The website to which you linked doesn't seem to understand the purpose and results of Prof. Schwab's experiment. In fact, it didn't really describe the experiment at all. It just rehashed a lot of quantum mumbo-jumbo to make it look as though some power of "mind" causes quantum effects, rather than the more mundane cause-and-effect of having to use tools ...


5

There are two parts to this question. Part 1: will the card slide?if I have a card at an angle, is there a limiting vertical force that will make it slide sideways? The force diagram looks something like this: This is a bit like the "climbing a sliding ladder" problem, in which case there is going to be a limiting force F - once you exceed that force, ...


5

When you inhale you create an area of low pressure immediately in front of your mouth, like the Venturi of a carburetor. You would be drawn toward the low pressure area as the incoming stream of air accelerates down your throat, maintaining the low pressure in front of your mouth. Until your lungs are full. When you turn 180 degrees and exhale, you reverse ...


4

In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ...


4

If you suppose that the scale works like a spring, which seems reasonable, then during standard use, the displacement $x$ of the scale is proportional to the mass $m$. The equilibrium relation is $$ mg=kx,\tag{1}$$ where $k$ is the stiffness of the spring. Assuming that, when you dropped a mass $M$ from a height $h$, all kinetic energy (which is equal to ...


4

Integrate the jerk 3 times then using starting conditions to work out the integration constants.


3

If you look at the instantaneous motion of your arm at any moment while it is flexing it will have a single direction; your arm goes up and then toward your center. During this time the opposite reaction is on the rest of your body, your torso is pulled slightly downward while you lift your arm and then slightly outward as you pull your arm toward your ...


3

Fundamentally, this is no different from computing the friction in a fluid (shear viscosity). The theory of viscosity goes back to Maxwell and Boltzmann, and microscopic calculations are possible for many fluids. Solid friction is more complicated, because the exact preparation of the surface obviously matters. First principles theories therefore concentrate ...


3

Newton's second law is a generalization of experience. It has no derivation in simpler terms.


3

The rotation period $T$ is given by $$T=2\pi \sqrt{\dfrac{a^3}{G(M_\text{Sun}+M_\text{planet})}}$$ where $a$ is the sum of the half axes of the ellipse. Routhly: $M_\text{Sun}=2\times 10^{30}$ kg $M_\text{Earth}=6\times 10^{24}$ kg $M_\text{Jupiter}=2\times 10^{27}$ kg If you assume both Earth and Jupiter are orbiting around the Sun (and neglect the ...


3

I was wondering that if they were orbiting in same orbit then will they both have same time period? If yes, then why because as they both have different angular momentum and both have so much of differences. I'll break this down into two parts, first looking at the period of individual objects orbiting the Sun at a distance of one astronomical unit (but ...


3

If the wire used around the pulley is considered ideal and massless, tension in the wire is same at each point and hence vector A and vector C are equal in magnitude. This reduces one variable and you will be left with 3 equations and 3 variables.


3

The essence of static friction is that it acts to prevent motion even in the presence of some outside force. The desk I'm sitting at while I type this is homemade and thus almost certainly not perfectly level. Yet the items on the desk are all fixed in place, not sliding down the slight slope. (OK, pencils tend to migrate by rolling, but..). It's the ...


3

I cannot see the image for some reason, but I think $\gamma$ is rather small there. The term $\gamma^2\omega^2$ shifts the maximum position, as a matter of fact. You took a rather strong "friction" ($\gamma=1$), which makes the resonance "frequency" smaller (longer period T). It is physically comprehensible.


3

We consider the integral: $$\sum_{i\lt j}\int_{t_0}^{t_f} \mathbf{F}_{ij}(\mathbf{r}_j(t))\cdot(\mathbf{r}'_j(t) - \mathbf{r}'_i(t))dt$$ For a rigid body, the distance between any two masses is always held constant, a fact that we can express as: $$\vert\mathbf{r}_i(t) - \mathbf{r}_j(t)\vert^2 = \Delta_{ij}$$ or $$(\mathbf{r}_i(t) - ...


3

Looking very close at the surfaces that touch at friction, this is an illustration Both surfaces are rough. They have ticks, holes, gabs, pits, spikes, and edges on the microscopic level. The smoother, the lower the coefficient of friction $\mu$. This constant is thus to be considered as a combined "roughness" between these two surfaces. Intuitively and ...


3

This is the simplest analogy I could think of. Imagine a long narrow carpet sliding across a huge ice rink at 1kph. On the rear end of the carpet stands a very fat (200kg) man wearing roller skates. You want to bring him to a standstill. You could grab the man and dig your ice skates into the ice until he eventually stops. Alternatively, you could grab the ...


3

You're mistaken. If I visualize , the speed of rotation increases as the person's hands fold inwards, this indicates that the angular velocity increases. The net angular momentum however, remains conserved. (If we consider the system to be isolated, that is. For every isolated system, the angular and linear momentum is always conserved.) Also, for the given ...


2

Nothing in physics can be proven in the mathematical sense. Sometimes mathematical physicists begin with axioms that are thought to model an aspect of the natural world and then work out what follows from these axioms. Their derivations are proofs in the mathematical sense of what follows from the axioms, but ultimately they do not prove anything about the ...


2

...a truck in motion and it has stack of hay (lets suppose) on the back. Now if the truck comes to a sudden stop will it stop faster if the force exerted by the truck on hay had overcome the friction force (another wording: will it be faster if the hay slips forward) or will it stop faster if the hay remains constant. I tried to find a braking ...


2

Why doesn't my hand just produce normal force on the book, cancelling out the force of gravity It does. Since the book doesn't accelerate downwards, another force is compensating for the weight. That is the normal force from your hand. and costing me no effort whatsoever? You are right that the normal force does not require energy to withstand the ...


2

Here's how. The ball gains a velocity $v$ due to gravity before hitting the ground. So each time it hits the ground its velocity is changed from $v$ to $-v$ (taking down as positive) during the collision, then returning again with $v$. The force $F_1= m\frac {dv}{dt}$ is experienced by the ball due to the collision, however this force is felt after the ...


2

All you need is 1.) the formula for the velocity of a falling body $$v_C = \sqrt{2g\Delta h} $$ 2.) the formula for the equilibrium in orbital velocity is $$v_C = \sqrt{\frac{GM}{r}} \rightarrow \sqrt{gr} $$ Just set $x = h_C$ and solve the system Inside the cart there are spring scales on which the person is sitting. His limbs are in the air ...


2

A collapsing gas cloud is an open system. It loses mass, energy and angular momentum as it collapses. Even if the net angular momentum of the cloud is zero, after the collapse the final planetary disk can have a significant net angular momentum, and the ejected material will have the opposite angular momentum. What can not happen, and that's where your ...


2

The issue here is that your front wheels are turned/steered by the same angle. When you try to find the instantaneous centre of curvature, you may first want to assume the wheels won't slip from side to side, like you may get if you drive around a corner on a slippy road. As there is no slip, the velocity of each wheel must occur in the direction the ...


2

From "what is a kettlebell" website: So just what is a kettlebell? A kettlebell is a cast iron ball with a handle attached to the top of it (picture a cannonball with a handle on the top). This design makes kettlebells different from training with dumbbells because the weight of a kettlebell is not distributed evenly, thus creating the need to counter ...


2

Okay, the system is in equilibrium so all the forces must be balanced. First consider the weights We have $W$ downwards from the centre of each of the 3 balls (assuming their centres of mass are at their geometric centres). We also have $N_A = N_B = \frac{3}{2}W$ upwards from the points of contact between the plane and balls $A$ and $B$ respectively. ...



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