# Tag Info

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Without providing a numerical answer to your question, to do so would help nobody: It would help to first draw a 'free-body-diagram', doing so will help you visualise the forces acting on the pole. You need to calculate the moment at the hand 1m from the end, this is your pivot, from the force exerted by the mass of the pole. Calculate the force required ...

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Force a alone will produce both downward motion (as if the force were applied at the center of mass) and rotational motion, because it causes torque. But the linear acceleration is the same, regardeless of where you apply the force. For traslation, both c and d balance a, so the disk will not translate in either case, but it will rotate with d because there ...

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All your thinking is very good and correct. BUT you are missing a point in the question :) Yes, the door's weight (which pulls from the center og gravity - no need to think about each particle of the door) creates a torque. The hinges then gives a counter torque. True. BUT let's read the question: A door is hinged at one end and is free to rotate about ...

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The torque equation is $$\tau = (r \times F) = rF \sin \theta$$ $r$ is a vector from the point from which torque is measured to the point where force is applied. $\times$ is the cross product $\theta$ is the angle between the position vector and force. F is the amount of force directed perpendicularly to the position of the particle. Any force directed ...

2

The discussion is mostly semantic. They are both calculated relative to a point, in the case of the torque the point has the additional meaning that if you put an axle trough the point, the object will start to rotatte around it if the net torque is not zero. It happens also that the torque will be the same if you chose any other point along the axis ...

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There can be no torque produced due to weight as they both are perpendicular to each other and therefore the work done by Gravitational force will be $zero$.

2

The tension in the rope should be different on the left than on the right - it is this difference that gives rise to the torque that accelerates the pulley. You seem to think that it should be the same: but if it was, then where would the torque to move the pulley come from? Annotate your diagram carefully: you did not show $T$ anywhere.

2

Your mistake is that the two tensions are different, because of the presence of a pulley with non-zero $I$. What you have missed is: 1) connect the two tensions to the torque: $(T_1-T_2)R=I\alpha$ and 2) link the accelerations $R\alpha=a_1=a_2$ NOTE (from comments): If the pulley had a zero $I$ (moment of inertia), then the two tensions would be be ...

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My analysis is that the object would not rotate at any velocity. My reasoning is that if the object is completely flat on the floor then there would be no net torque. Say r is the distance from the bottom left corner (as in your diagram) to the point at which the net frictional force acts (directly beneath center of mass, at a point where the object and ...

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Because: $$\vec p = \vec r_2 - \vec r_1 -r_3,$$ where $\vec r_3$ is the unlabelled side of the triangle that is parallel to $\vec H$. $$\vec r_2 - \vec r_1 = \vec p + \vec r_3$$ $$(\vec r_2 - \vec r_1) \times \vec H = \vec p \times \vec H + \vec r_3 \times \vec H$$ but $\vec H$ and $\vec r_3$ are parallel so $\vec r_3 \times \vec H = 0$, and $$(\vec r_2 ... 1 Here is the FBD as I understand the question The balance of forces along the incline give us$$ C - W \sin \theta = 0 $$The balance of forces along the normal give us$$ A + B - W \cos \theta = 0 $$And the balance of torques about the bottom contact point are$$ \ell B + h W \sin\theta - \frac{\ell}{2} W \cos\theta = 0  All this works out if you ...

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Torque is not a force. You can say there is a torque caused by normal forces, but there is no special name for that. A normal force comes from acting with a force on an object resting next to a surface. The surface prevents the object from moving through it by producing a reaction force that is necessarily normal (perpendicular) to the surface (parallel ...

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Because you have to push it further.

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