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Here is my derivation of this result. I hope you find it helpful: Say we have n different forces $F_1, F_2, F_3... F_n$, applied at n different points. Now we pick two centers $P$ and $Q$, and express the radial vectors (1) from point $P$ to each of the n points (where forces are applied) as $r_1, r_2, ... r_n$ (2) from point $Q$ to each of the n points ...

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The 2nd solution you wrote down appears to be the correct solution. Offhand, I see two problems in the first solution. First, I think that a problem with the first attempted solution is that you made a subtle mistake in assuming that F=ma means that $F=mr_1α$. That seems like a plausible step at first but if you examine this step more closely you'll realize ...

0

This analysis assumes one of two things. No acceleration If the beam is rotating at a constant velocity, then the fact that there is no net torque on the beam is fine as the net torque is equal to teh moment of inertia times the rotational acceleration $\tau=I\,\alpha$ No mass The net torque could also be zero and still allow for acceleration if the mass ...

1

If the lift angle is $\theta$ (shown at zero in the diagram) then the payload lever arm is $$x_1 = \tfrac{L_1}{2} \cos \theta+L_2 \sin\theta$$ The force lever arm is $$x_3 = L_3 \cos\theta$$ Static balance exists when $$\left. \vphantom{\int } (M g) x_1 = F x_3 \right\} \\F = \frac{x_1}{x_3} M g = \frac{\tfrac{L_1}{2} \cos \theta+L_2 \sin\theta}{L_3 ... 1 You need to balance the moments about point P. The horizontal force F time L_3 will equal the mass M times the horizontal distance L_4 between P and M.$$F \times L_3 = M \times L_4 This calculates the force required in the current position you've shown. Worst-case, an infinite force will be required after rotating $90^{\circ}$ from that shown, ...

3

On a well designed balance scale, the center of gravity of the beam or lever arm will be just slightly below the center pivot point. If the beam is not level, the center of gravity will be to one side or the other of the pivot, and will thus create torque as it tries to move directly below the pivot point. The distance between the pivot point and the ...

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The problem is when you say "There is no net external torque because the end of the rod is accelerating linearly along with the center of mass." That's not how torque works; torque doesn't care about the net acceleration. The torque is only a function of where you apply the force and in what direction. In such a case as you have described, the $\vec r\times ... 0 Your object has a center of mass, and for most practical purposes you can treat all of the mass being centered at that point. Mass has inertia and resists force by Newton's 2nd law as you have written. But If you apply the force not through the center of mass, but rather away from it you now have to treat the mass in a distributed manner. The mass at the ... 0 Forces and torques (or moments) are vectors. In a nice coordinate system, like the mutually perpendicular x,y,z axes we usually use, all possible forces and torques can be represented as a combination of forces or torques along the coordinate axes. This, basically, is why we use a specific coordinate system with basis vectors. For example, any torque along ... 1 If, as shown in the diagram, the main rotor is moving in the counter-clockwise direction, then the body of the helicopter will try to twist in the opposite direction (i.e., it will want to turn clockwise). The tail rotor thus needs to provide a thrust which gives a counter-clockwise torque in order to counter and cancel the clockwise torque on the helicopter ... 1 The short answer is that the 770 HP is the more powerful. However, I suspect the question is about the difference between power and torque. 'Torque' is the rotational equivalent of force and if an engine is exerting a 'high torque', it's pushing hard. This occurs when the vehicle is accelerating, especially from rest. 'Power' is the rate at which ... -1 The 770 horsepower one. The torque just refers to the ability of the engine to turn the shaft, which is then converted by the transmission. However, the horsepower refers to the actual amount of power or 'acceleration' the engine can dish out. It is the job of the transmission to ensure that the engine has the ability to efficiently supply power to the ... 0 Here author consider two extra oppositely directed forces along the rod for the seek of calulation. This two forces being oppositely directed, there will be no net force for this and would not cause any rotation of the rod. Now the combination force of$f$and$F_2$gives the resultant${F_2}^{'}$which makes an angel$\theta_1\$ with the rod. similarly the ...

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