# Tag Info

29

The problem of perception as to "What is a new Equation of Motion?" seems to originate with the dogmatic teaching of the Three Equations Of Motion as a Set of Results to be Learned. They are in fact three results derived from the distillation of Newton's Laws: $$\mathbf f = \dfrac {\mathrm d} {\mathrm d t} (m \mathbf v)$$ which differential ...

28

There's nothing wrong with your calculation. But plug in $v=u+at$ and you find: $$x=(u+at)t-\frac{at^2}{2} = ut + \frac{at^2}{2}$$ which is already one of your three equations of motion. So your teacher is probably just saying that it is redundant to call this a "fourth" equation, because it is essentially the same as ...

20

The ball will slow down to terminal velocity. This is because the force of air drag increases with increasing speed. Terminal velocity is the speed where the force of air drag equals the force of gravity, so the total force is zero and the object travels at a constant speed. If the ball has a higher speed, then it will have an air drag force greater than ...

19

It does. The fact that it is in contact with the earth (or rather, the rails), just means that it cannot move sideways under the force, because the rail restrict its movement. In the same way, if a train travels down a rail and has wind blowing from one side, there is still a force pushing it, but the rails will exert an opposite force so it can't be moved ...

11

If the centre of gravity of the object is vertically above the edge of the table then the object is in equilibrium. However, this equilibrium position is unstable (like a pencil balanced on its point) because a small tilt of the object will lower the centre of gravity, which will then cause the tilt to increase. This is a positive feedback loop. However, if ...

7

Just to add that when you use these equations it's good to think in a physical way as you go along. So: $$v = u + a t$$ think: "yes the final velocity is the initial velocity plus the change owing to acceleration. This is basically the definition of $a$." Next $$x = u t + \frac{1}{2} a t^2$$ think: "the distance is travelled is how far it ...

5

Philosophical aspect Physics typically cannot answer "why"-questions very well. Physics observes phenomena in the world, creates descriptions of the behaviour observed there, and postulates rules (physics laws) stating that under specific circumstances this very same behaviour will happen again. So, the best answer that physics can give go along ...

5

Soft sand deforms when you step on it. The force applied by your foot to the sand moves through the deformation distance, performing work on the sand which your muscles provide. That work does not assist your propulsion; it is lost to the sand. The sleek flat floor does not deform when you walk on it and therefore does not "steal" work from you as ...

5

If you recall the Coriolis force given by $$\mathbf{F}=-2m\mathbf{\Omega} \times \mathbf{v}$$ If for a minute, We don't get into the detail of the analysis but just see the magnitude of this. $$|\mathbf{F}|=2m|\mathbf{\Omega}_\text{earth}||\mathbf{v}_\text{train}|$$ Let's take the speed of the train to be $100$ Km/Hour which is about $30$ m/sec and mass to ...

5

Well, few problems here, which all seem to come down to a disregard for units of measurement. First, the quantities you take to be equal to $1$ in the beginning still have dimensions (apart from the drag coefficient, let's keep that at $1$ for simplicity), so they're not just "$1$". In fact, you see that shortly after this, you're writting results ...

4

Terminal velocity is achieved when the drag force $f=Dv^2$ (where $D$ is a constant and $v$ is the object's speed) equals the gravitational force $F_g=mg$. Equating these forces is the condition for terminal speed, which is $\displaystyle v_T=\sqrt{\dfrac{mg}{D}}$. If the ball is thrown with an initial velocity greater than the terminal velocity, then $f>... 4 If you approach this problem mathematically and tried to calculate the time it takes for an object to tip over as a function of the overhang distance$x$of the center of mass over the edge, you will find that$t \rightarrow \infty$as$x \rightarrow 0$. So does this mean that the object will tip but after an infinite time, or that it won't tip at all. ... 4 By resolving it's doesn't mean that it's sort of resultant of$u$and$w$because$uw$are used as axis. You should say one can write$\mathbf{F}$as linear combination of component along$u$and$w$so that $$\mathbf{F}=F_u\hat{u}+F_w\hat{w}$$ where$\hat{u}$and$\hat{w}$are unit vector along these axis. 3 Neither formula is exact (assuming that Newtonian gravity is the only force affecting the two masses): a reference frame in which$m$moves circularly while$M$is stationary is not inertial, since in an inertial frame$M$will move by the force exerted by$m$. The same is true for the reverse. The first formula holds in the approximation that$M \gg m$: ... 3 Assume the object hangs vertically before the bus starts to accelerate. When the bus starts to accelerate there is instantaneous relative movement of the object relative to the bus seen by observers both on the bus and on the ground. To the observer on the bus, the bus is not moving and the object moves due to the fictitious force present in the accelerating ... 3 What is the value of acceleration when the string breaks? The string will break when the tension in string reaches the tensile limit say ($T_0$).The tension in string for given acceleration can be calculated easily from FBDs. Why does not the ball stays fixed in its initial position forever, so that the tension becomes high in the string, and it eventually ... 3 No, wings are designed to carry the maximum load with a minimum of weight. This results in their being flexible; to make them perfectly rigid would dramatically increase their weight and thereby reduce the payload of the aircraft. Aerodynamic flutter resulting from wing or control surface flexibility is then managed in a way consistent with minimizing the ... 3 The modulus of elasticity (Young's Modulus) is defined as stress divided by strain in the linear elastic region of the stress strain relation for a material, or $$E=\frac{\sigma}{\epsilon}$$ Where$\sigma$equals the stress. The units for stress can be in various forms. Non SI (engish) units for stress are typically PSI (pounds per square inch). The SI unit ... 3 First of I think it is great that your taking apart equations and putting them back together in a different ways as if they were LEGO pieces. I think most of us (with an interest in physics) did that in high school and it can be very insightful. So instead of fretting over the characterization of$x = v t - \tfrac{1}{2} a t^2$, debating if it is an equation ... 3 No, we cannot. In fact, the claim is not even true. Bodies do not fall with constant acceleration. The gravitational force decreases with the distance squared, so as the body falls its acceleration increases, if only a bit. Plus, there is air friction, etc. If the falling distance is short enough that you can neglect the dependence of the gravitational force ... 2 When I learned the kinematic equations in high school, this equation was actually presented to us. There is however a reason it's seldom seen. The main "issue" is that your equation, in order to be valid for all$t$, would be better expressed as $$x(t)=v(t)\cdot t-\dfrac12 at^2$$ This is because when$t$changes,$v$changes as well. And if you ... 2 Your first approach is wrong. You assume that the work is$F\cdot S$, which is only valid iff$F$is a constant force. However, the force that satisfies your scenario is not constant. The easiest way to show this would be take the negative gradient of the potential energy which gives the force. If you haven't learned this, then you can try solving for the ... 2 One of my mentors likes to say, “Never make a measurement unless you already know the answer.” By which he meant that, any time you are building a new measurement instrument, which is what most experimental apparatus is, you’re going to discover several times that what it’s measuring isn’t quite the same as what you thought it was measuring. Here you have a ... 2 One thing that hasn't been raised is that your derivation has a mathematical problem in it. In the step: $$(v+u)(v-u) = 2 \frac{(v-u)}{t}x$$ $$(v+u)t = 2x$$ you have divided each side by $$(v-u)$$ The problem with this step is that this could be the$0$, unless you explicitly require $$v \neq u$$ Dividing by$0$is not a valid operation. It can be ... 2 Elastic moduli are energy densities -- they have units energy/volume. 2 Suppose you have one spring of equilibrium length$l$and constant$k$. A thought experiment you may conduct is to consider it made of two springs of length$\frac{l}{2}$and constant$k'$. Let's try to determine$k'$. Well, by the equation which describes series associations of springs, you should have$\frac{1}{k}=\frac{1}{k'}+\frac{1}{k'} \Rightarrow k'=...

2

I will give my answer, but I urge you to take a quick look at the code of conduct before asking future questions. Firstly, it does not help us at all for you to overstate your background. We're not here to judge you, it's okay to admit your lack of knowledge, that is what this entire website is for. And most importantly it helps us to answer your question in ...

2

For example, if an object stays still at 15°, but moves at a constant velocity at 20°, the kinetic coefficient of friction will be greater than the static coefficient of friction. How does that make sense? If the object is still at 15$^0$ all you know is the static friction force up the incline equals the force down the incline. The static friction force ...

2

Since the two masses aren't connected via friction or using another medium (such as a lead or a spring), hence you can't consider the two masses as a unique mass (or as a system of connected body). Every mass has to be considered separate from the other ones. So the equation that holds for this situation is, as you wrote: $$\vec{F} = M_1 \vec{a}$$ which you ...

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