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Actually with the lever sticking up above the balance scale shown; balance axis point; it adds gravitational weight to whichever side is initially positioned downward, and also since this side is down, it should weigh more being closer to the ground... and should tend to keep this side down... However is doesn't... If you put a weight, on some point of a ...


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The problem with attempting to do the analysis with the forward point of contact on the box when it is sliding is that the box is accelerating. This makes a non-inertial frame and there's more moving parts. Besides the force of gravity on the center of mass, there will be fictitious forces. First, lets assume friction is zero. If so, we can calculate ...


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Take a look on the forces diagram in the first picture. Let's ignore the force $N$ for now, it can only contribute to the tipping (due to its orientation), but since it is going to be confined to acting only on the red point as soon as the tipping occurs (if it does), it won't have any influence at all. E.g. it produces zero torque about the red point. ...


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It's part and parcel of the definitions of angular momentum and torque. Consider a system of particles, not necessarily a rigid body. Without loss of generality, we can use an inertial frame that is instantaneously co-located and co-moving with the system center of mass. The angular momentum of the system with respect to the origin of this frame is defined ...


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The physical explanation for why torque increases with r is that the longer the lever arm is the greater angular acceleration you can cause for a given force F. If a screw is stuck because it was screwed in too hard (ie with too much torque), you need to get a longer wrench. With the longer wrench (ie, larger r_w) you can generate greater force at the edge ...


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Why would there be no radius in torque? A real torque is a real force that acts on a real rotating body (rigid or not) at a real radius. If you look at e.g. rotating machine parts, they all have a finite diameter. That diameter is of enormous importance for the design of a part, because together with the material constants it determines just how much torque ...


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Can a ball stay still while laying on a inclined plane? In freshmen physics, the inclined plane and ball are perfect and the ball moves, so for your purposes, no. If either the surface or the ball have imperfections, we can tip the plane and the ball won't move until gravity exceeds the sum of the normal forces. To imagine those normal forces, we look ...


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General remarks. That's right. Torque is defined as $\mathbf r\times\mathbf F$ where $\mathbf r$ is the position where the force is applied, and $\mathbf F$ is the force being applied. The so-called Law of the Lever can then be derived from the following fact (which itself can be derived from Newton's Laws) about systems of particles: The net external ...


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As you say, torque can be defined as $ r \times F $. Then the law of the lever is derived simply from the condition that at static equilibrium all forces on the lever and all torques about the fulcrum sum to zero.


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If you are just interested in the rate of precession, then $r$ just has to be the distance from the support (tip of the top) to the center of mass - this is the distance that gives rise to the torque through $mgr\sin\phi$ However, if you want to do this as a vector equation, then you have the instantaneous angular momentum (vector) of the top $I\vec\omega$ ...


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To remember better how it works, you can just tell yourself that units follow the same operations as numbers. You obviously knew you had to do a product to get "30", now do the same with units. (as Danu is implying)


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You should think about the definition of torque: $$ \vec \tau = \vec{r}\times \vec F\implies \tau =rF\sin\theta$$ where $\theta$ is the angle between $\vec r$ and $\vec F$, the second equation refers to the magnitude of all quantities. If the units of $F$ are Newtons, and the units of $r$ are meters, then what should the units of $\tau$ be? Remember that ...


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Newton's second law $\vec F_\text{net}=d\vec p/dt$, or the-more-convenient-for-simulations $d\vec v=(\vec F_\text{net}/m)\,dt$, still applies. It doesn't matter where the force is applied on the body; the instantaneous acceleration of the center of mass will be the same as if the forces were applied at the center of mass. As for your pen example, try doing ...


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The angular velocity of the car equals the tangential velocity of the rear wheels divided by the radius of turn. $$\omega = \frac{v}{\rho}$$ The radius of turn is found from the steering geometry, but in general you can simplify it by $$ \rho = \frac{ \ell }{ \tan \theta} $$ where $\ell$ is the wheelbase of the car and $\theta$ is the steering angle. ...


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Friction does not depend on velocity (unlike viscous drag). An object that is stationary on a table will continue to be stationary when you push it gently - because there is an opposing force of friction. So no, your understanding is wrong: friction is present even when the object is just starting to move. Let me draw a diagram: That ought to clear it ...



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