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5

You see, when you have a pendulum with friction you account it by including a force $\vec{F}_r=-b\vec{v}$. Then your differential equation for the pendulum is $$ml\ddot{\theta}=-mg\theta-bl\dot{\theta}\iff\ddot{\theta}+\frac{b}{m}\dot{\theta}+\frac{g}{l}\theta=0$$The solution of this differential equation depends on the values of $b$, $m$ and $l$ but since ...


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The contact point between the wheel and the car is stationary - there is no "rubbing" there (well there is because the contact point is really a patch but let's keep it simple). To do work you need "force times distance" - and without relative motion there is no "distance". When you apply the brakes in a car you have sliding of the pads relative to the ...


2

I think you are confused about what $d$ is supposed to mean in the equation $W=F\cdot d.$ You seem to be under the impression that $d$ is the distance that the object being acted on moves relative to the object providing the force. But this is not the correct meaning of $d$ in the equation and you know it. Imagine if the car crate were in front of the ...


1

The correct version: $$P \cos\theta-F_{fr}=ma.$$ Here $P\cos\theta$ is a projection of $P$ on the direction of motion. Next we use the expression for the friction force $F=\mu N=\mu mg$ (note that $N=mg$ in this case because the motion is happening on the horizontal plane). Now $$P \cos\theta-\mu mg=ma.$$ From where we find $$P=\frac{m(\mu ...


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By definition, the work done by a force is $W = F\cdot d$, so how can static friction do work ? Can this force move the body a distance of $75~\text{m}$ ? Friction does negative work on the truck, slowing it down and does not move it forward. What does positive work on the truck, accelerates it and makes it translate $75~\text{m}$ is the engine ...


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I am reasoning my way to an answer here - I have never seen this problem before so I could be completely wrong. I think the issue is that you need to make sure the normal force is reduced as the tension on the "signal" side of the capstan is reduced. This is presumably why the slightly stiffer wire (as opposed to the rope) gave more predictable results at ...


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I don't think that equation is right. $F_{max} = \mu mg$, so $a_{max} = \mu g$. Where are you getting velocity from? The spring doesn't move at a constant velocity, does it? You need to use the spring's maximum torque and work out how to weaken it so the final acceleration is sufficiently low. Why are you trying to make it as fast as possible? If you're ...


1

How can it require less petrol to run the engine at 2,500rpm pushing a heavy car at 55mph than the exact same engine doing the same rpm only pushing at 20mph (far less air resistance)? It doesn't require less petrol, it requires more. However, if it requires less than $\frac{55}{20}$ times more petrol at 55mph, then the car is more efficient ...


1

If the ring is stable it is in equilibrium even if it is not in the center of your force table. In that case, while the forces will be given by the $mg$ values, the angles that you are reading are not the actual vector angles. The ring will move so that the forces balance out. The angles you read on the edge of the table are correct only if the ring is ...


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$m_1 = m = 10$ kg $m_2 = 2m_1 = 20$ kg $F = 60$ N; $ma = F - \mu mg-T$ $2ma = T - 2\mu m g$ $\Rightarrow ma+\mu mg = F-T$ $ 2(m a+ \mu mg)=T$ $\Rightarrow 1/2=(F-T)/T \Rightarrow T = 2F/3$


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As the wheels try to roll they are prevented from rolling by the frictional force acting in the opp. direction.As the traction force exceeds the limiting frictional force the wheel starts rolling forward w.r.t rails. The force tries to induce relative motion between the wheels and rails.As the rails cannot move backwards (due to friction) the wheels have to ...


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When you have a lightly damped oscillator, there is a small correction to the resonant frequency. This is derived in detail on the wiki page for the harmonic oscillator. The form they give is $$\omega = \omega_0\sqrt{1 - \zeta^2}$$ Where the $Q$ (quality factor) of the oscillator is given by $Q=\frac{1}{2\zeta}$.


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The force due to sliding friction causes the center of mass to decelerate - since it is the only external force acting on the ring, and we know the center of mass moves as though all forces act there. You might find this earlier answer of mine and the links therein helpful. As for the "in the real world the ring will stop completely" part of your question - ...



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