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Let us consider there is no external force and the tyres are rolling smoothly without slipping. Here friction doesn't come into play let me explain you how. Even if friction is present, this friction is not the answer. The concept of friction most books provide are deficient. The term "rolling friction" is also a misnomer. The correct answer is rolling ...

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The tyres of the cycle are rolling and the remaining cycle moves with a velocity same as that the centre of mass of the tyres have. Now the question is which force is responsible to bring the cycle at rest. The answer is Air-friction and Rolling-friction. It should be noted that the static and kinetic friction does not come into the picture because the point ...

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You are thinking about energy, and not about force. The charge exerts a force upon the system, or the system exerts a force upon the charge. Energy is consumed in that process, but that isn't the point of the statement made. Look at it through Energy $E$.(the field will be $\mathbf E$) Lets look at these formula: \begin{align} E&=\int_\ell \mathbf F ...

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To move charge from one point to another in an electric field, the force which we must apply is equal and opposite to the force due to the field. The sentence you provided is actually confusing. I think it should have been: To move charge from one point (lower potential) to another (higher potential) in an electric field, the constant force which ...

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Remember that, if the net force is zero, velocity is constant (not necessarily zero!). You only need to do push a tiny bit harder for a tiny bit of time to start moving the charge. This extra amount can safely be ignored. Once the charge is moving with some nonzero velocity, equal force is enough.

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What I was doing wrong was putting manufacturer-provided motor torque directly into the formulas; actually the overall gear ratio must be taken into account, and it results from data taken around on internet that for electric vehicles it is =~8 (dimensionless). Hence the proper expression to use for $v_f tanh(\frac{F}{mv_f} t)$ is not $v_f ... 1 The answer to this question depends on whether you're working "in real life" or on a physics problem. On a physics problem, the object doesn't start to move. The net force is exactly 0 at this time, so there is no acceleration. In reality, however, the object would possibly move. There's a variety of reasons for this: The normal force, static ... -1 The inverse tangent of the coefficient of static friction is$17^{\circ}$. 0 Well, I'm not sure if you can use the ideal friction for this case. But the surface area does matter, if you want to light something up. Given the same normal force over 2 different surface areas, you would experience the same resistive force right? So given the same moving velocity (simplification), the power dissipation should be the same. The question ... 0 There are a number of wrong assertions in your statement. You say, "surely ... continue accelerating to infinity." Since$ a = \frac {F}{m} $, as long as you apply net force F, you only get constant acceleration a. The acceleration value not only does not go to infinity, but actually does not change, unless the net force changes. When there is no net ... 0 Since$m_1$is required to take the largest possible value, therefore we should consider the case when$m_1$just begins to slip down the incline. In the condition just before the slipping begins(still a state of rest), the equations of motion should be:$m_1g \sin \theta-T-f_s=0\ \ \ \ \ (where\ \theta=35^o, \ and f_s\ is\ the\ force\ of\ static\ ...

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The answers posted by others encouraged me to examine the assumptions I was making in trying to solve the problem. Unfortunately I think none of the answers were ultimately heading in the right direction so I can't accept any of them. I believe the answer is that the friction between block B and the plane will always be at its maximum value once the plane ...

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$f_{AB}(t)$ will be on the left side and $f_{BF}(t)$ on the right. Maximum $f_{BF}(t)=\mu (m_A+m_B)g=(0.6)(30)(10)=180N$ So, $F(t)$ will have to be greater than $180N$ so that $B$ can move. When it does, $A$ will experience a pseudo force say $F'(t)$ in the left direction. If $B$ accelerates at $a_B$, then $$F'(t)=m_Aa_B=10(180/30)=60N$$ But, Maximum ...

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I am adding another answer in order to show the steps needed to solve such problems in the general sense. An acceleration run is split into parts of differing acceleration domains. Given a starting condition for time, distance, speed, acceleration of $t_0, x_0, v_0, a_0$ here is what you can do Constant Torque with Air Drag $t=t_0 \rightarrow t_1$, $v=v_0 ... 0 There are two things that limit the maximum traction (F) of a car. One is given by the friction formula, F = μR (Traction = friction coefficient x weight of car), above which the wheels start to spin. The other is the power equation. P = Fv or F = P / v (Traction = power / velocity), which the engine isn't powerful enough to exceed. Note that the max ... 3 What you need for circular motion is Centripetal Force. Definition: Centripetal force is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion. If the road is ... 2 You have to split the time domains into the gears needed to reach 60mph. For each gear, there have to be assumptions on the power delivery of the car. Typically 1st gear is traction limited, so you can assume constant acceleration up to the speed where peak power occurs. The relationship between power speed and acceleration is$P(v) = m \,v\, a(v)$. So run ... 0 Here, because the coin is placed at the center, the centrifugal forces balance each other. Every point mass in the coin has it's conjugate point at the diameter passing through it and on the same distance from the center on the other side. Hence the coin is under equilibrium and does not fly off. 0 The coin will not move. First, to differentiate between centrifugal and centripetal, I'll start by stating the definitions first. Centrifugal force is the apparent force that draws a rotating body away from the center of rotation. It is caused by the inertia of the body as the body's path is continually redirected. Centripetal force is a force that makes ... 1 The kinetic enegy at$t=0$is equal to the friction work done when the wheel has stopped:$E_{kin}=W_f$that is$1/2J \omega_0^2=F_fs=mg \mu 2 \pi r n $(with$J=mk^2$) where$r$is the radius of the bearing bore an n the number of revolutions. Solving for the revolutions gives:$n= \frac{J \omega_0^2}{4 \pi m g \mu r}$The angular acceleration due to ... 0 I think Sir Issac Newton's laws works fine here. 1st law Object will continue to be in rest or move in straight line until a external force applied on it. the ball starts/stops moving only when an external force applied on it. 2nd law F=ma The ball uses the force 'F' applied on it to move in acceleration 'a' on direction of force. 3rd law Every ... 0 The answer is that there is friction. Friction is a dissipative force, meaning that friction will cause energy losses. This causes the ball to stop rolling: in a vacuum that would not happen! These things can be taken into account when doing calculations. 3 You are talking about Newton's first law, i.e, inertia (ball should keep moving) This has absolutely nothing to do with the current group-theory tag. The first law doesn't need to explicitly mention friction, it already says "when no force is applied". It's upto you to keep track of external forces. This has been voted for migration to Physics but I doubt ... 1 At low angles when friction can hold the blocks together, then there is no tension on the cable and thus: $$f_2 = m_A g \sin \theta \\ f_2 = (m_A+m_B) g \sin \theta$$ Only when there is motion on the blocks there is tension. In that case you have$\ddot{x} = \ddot{x}_A = -\ddot{x}_B$f_2 = \mu_S m_A g \cos \theta \\ f_2 = -\mu_S (m_A+m_B) g \cos ... 1 Water and air are both weakly magnetic, with magnetic permeabilities on the order of$10^{-6}$H/m whereas iron, a strongly magnetic material, has a permeability of about 0.25 H/m. Thus, if you could create the supercavitation in water, a magnetic field would do just about nothing to maintain the cavity. 1 Correction, the fourth equation should be$T+f_1+f_2=m_BgSin\theta$Tension will be produced in the string if the blocks are stretched from the other side. When$\theta \leq 21.8^\circ$, you calculated that the frictional force on$A$will be more than the force due to the incline. So yes,$A$will accelerate upwards, slacking the string, so no Tension will ... 1 Hint : Friction opposes tendency to move. Tension is produced if the string is stretched$very$slightly. So, increase friction to maximum and then tension will act if necessary. Ironically, you are thinking absolutely right. Give yourself a cookie. From part$a\$, we know that the blocks will be at rest at all angles below that. You are also right as at ...

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Sorry, but there's no obvious way to figure out this stuff a priori. It depends critically on things like how smooth the surfaces are. And it does not, to a first approximation, depend on surface area. About all you can tell about sliding vs static friction is that sliding friction is less than static.

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The reason why wider tires are chosen over thiner ones is not because of greater friction. And the reason is not complicated, or does not "start a complexity".. the reason why wider tires are chosen is because they last longer than thin ones! Nothing about friction at all. Racers would always have to replace thiner ones, so they figured, why not get larger ...

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First remember that it isn't the acceleration/force that matters. It's the velocity -- friction opposes motion, not acceleration. So if the block starts with a upward motion, the friction will point down, no matter what the forces are. So you definitely need to know about the direction of the velocity. If you have an initial velocity, this isn't a problem ...

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Since we know that the block is restricted to move only along the inclined plane, all the force components perpendicular to the inclined plane come in action-reaction pairs( assuming that the inclined plane is stationary and perfectly rigid ) and therefore I shall not include those components in my free body diagrams in order to make things clear. First, ...

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