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In your example you have the total load applied as $$F = \int \limits_0^\ell w\, {\rm d} x = w \ell$$ where $w$ is the linear force density (in Newtons per meter). To get the torque you just include the position of the force $$\tau = \int \limits_0^\ell x\, w\,{\rm d} x = \frac{1}{2} w \ell^2$$ The rule for distributed loading is to find the total ...

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In the case of the gate, F=200N is the distributed force, which is applied from x=a=0m to x=b=2m from the axis. The force applied on an element of length dx is (F/L)dx Newtons where L=b-a. The torque on the gate due to the force on this element is dT = (F/L)xdx. We integrate this from x=a=0m to x=b=2m to get the total torque : $T = [\frac{F}{L}\frac12x^2] ... 1 A force applied uniformly to a uniformly dense object (or with a uniform-force-per-unit-mass even if the object is not uniformly dense) is equivalent to the same total force applied at the center of mass of the object. The equivalence means the same total force and the same total torque. In particular, if the center of mass is chosen as the origin of the ... 0 You will find it significantly easier to solve the problem by using the conservation of energy. Remembering that for the no slipping condition it is the perpendicular distance of the centre of mass above the rails which is the important parameter for the no slipping condition. That perpendicular distance you need to relate to the radius of the sphere$r$and ... 0 You are right Jeff. Using a static number for torque, and no velocity dependent frictional term will give you an unreasonably increasing acceleration. As your intuition predicted, electric motors start out with high torque at rest and decrease to zero torque as speed increases to max speed (Good discussion here). According to your spec sheet, your motor has ... 0 I recommend using an accelerometer to find the experimental acceleration of the robot. Probably the best quality would be to use a LabQuest and/or LoggerPro paired with the Vernier Motion Sensor, but you could also use SparkFun accelerometers alongside an arduino/RPi. Another option, which very well may be the best given constraints for your time and ... 0 From the equations of static equilibrium, N - mg = 0 in the vertical direction. The tipping is caused by the imbalance in the moments applied around point P. As long as mg * (L / 2) > F * H, then the book case won't tip over. H and L are the dimensions of the book case, so you can't do much about changing them. You can either add more books, which ... 1 You almost solved it. Actually when you increase your force steadily the normal reaction shifts it point of action from directly under the centre of mass of the object to the point P. In other words in the limiting condition the normal force acts at point P. And hence it's torque is 0 about P. Well done. 1 One promising way of implementing arbitrary physics simulations, is by programming in terms of 'constraints'. I highly recommend reading this article that covers constraints very well: http://gamedevelopment.tutsplus.com/tutorials/simulate-tearable-cloth-and-ragdolls-with-simple-verlet-integration--gamedev-519 I was very surprised when I first saw this ... 1 The meaning you quote is only one of several. From the same dictionary, others which are now obsolete or not often used are 3: importance in influence or effect 4 obsolete : a cause or motive of action and it is from these that the scientific meanings derive : 6a : tendency or measure of tendency to produce motion especially about a ... 0 In the language of physics, a moment is a physical quantity which accounts for how a physical property is located or arranged. We do NOT use the colloquial meaning of the term moment. This "moment" we are referring to is derived from the latin word momentum which is also a physical quantity equal to the product of the mass and velocity. 0 The moment of a vector$\vec v$applied in the point$\vec p$with respect to the pole$\vec q$is by definition $$\vec M = (\vec p - \vec q) \times \vec v$$ With$\times$vector product. 0 You have used the correct definition of the torque due to a force about a point as the force times the perpendicular distance from the point to the line of action of the force which are you green (length$2a \sin \theta$) and yellow (length$4a sin \theta$) lines but not realised that there is a net torque about$O$on the system of$4mg a \sin \theta$... 0 Consider the figure above. The torque about$O$is given by $$\vec{T}=\vec{r}\times\vec{F}$$ Its magnitude is $$||\vec{T}||=||\vec{r}||.||\vec{F}||.\sin{\theta}$$ As you can see in the figure,$||\vec{r}||.\sin{\theta}$is equal to the length of the green line, which is the perpendicular drawn from$O$to the line of action of$\vec{F}$and is ... 0 The question is rather unclear: if the force under consideration is the weight$mg$then the lever arm is of course the yellow red, since it must be perpendicular to the force. 1 You have made a model of a viscous fluid coupling which was used in a number of four wheel drive vehicles to transfer torque. The system relies in the fact that adjacent planes of moving liquid experience a viscous force between them. 5 It is due to the viscous nature of any liquid. When you stir, the liquid starts spinning and this causes the liquid (the part which is in contact with the pot) to "drag" the pot(due to friction) along with it in the path of its motion.Hope this answers your question. 2 A probable explanation for this effect is simply that the bottom of the pot might be a bit bulged out, as to form only one point of contact around which the pot then can rotate relatively freely (with little friction). As you stir the water inside the pot, the moving water molecules exert a frictional force on the walls of the pot, dragging it in the same ... 0 The space ship should undergo both linear motion and angular (rotational motion). Linear motion will take place in the direction in which the force is applied, and rotation about its centre of gravity. So the centre of gravity will act as an imaginary pivot. 1 Hint: first find the angular velocity of the disc by conservation of energy,$\frac{1}{2}I\omega^2 = mgh$,$\omega\$ = angular velocity of the pulley, I = moment of inertia of the pulley. It is mentioned that the pendulum cover a horizontal distance L in t seconds. And we also know that $$v_{linear-velocity} = R\omega$$ So time t can be found ...

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When you don't have to consider the inertia of the screw (which is always), these problems can easily be analyzed by unwinding the screw to a plane (as you've done in your figure). So think of the screw and the surface of contact as two inclined planes that slide against each other (one on top and one beneath it). Now, pushing on the screw is like pushing ...

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Gareth, this is a good first question. You've asked it well, and I encourage you to ask more questions in the future. Basically, you can just reverse any usual explanation of a screw. Normally, the wedge transfers torque into thrust; well in this case the wedge transfers thrust into torque — it's exactly the same, but reversed. This is a special example ...

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