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Why is torque or moment of couple independent of choice of point of rotation? Torque is given by $\tau =\vec r\times\vec F$. **It depends on r **, which is its distance from axis of rotation to the point where force is applied. A couple is defined as: "Two parallel forces with the same magnitude but opposite in direction separated by a ...


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Your assertion that "torque or moment of couple independent of choice of point of rotation" is only true either (1) of a system of forces $\vec{F}_i$ at positions $\vec{R}_i$ that sum to nought or (2) you shift your torque computing centre along a vector parallel to any nonzero nett force. Otherwise your assertion is not true. The fundamental reason for ...


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This follows from definition of 'torque' and 'couple' and is a simple matter of geometry. In the diagram above, a couple is applied to a disk of diameter D. That is, a force F is applied to opposite sides of the disk. The torque do to the couple is: $\tau = F\times d_1 + F\times d_2$ where $d_1$ and $d_2$ are the distance to some (arbitrary) point, $O$. ...


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A net force on an object likes to act through the centre of mass of that object; in this case, the middle of the door. Apply a force to the centre of mass and it will accelerate uniformly because there is an even amount of matter on all opposing sides that can resist the motion with an even amount of inertia. If a force is applied off the centre of mass, it ...


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There are many variables that come into play in an armwrestling match, but one answer as to why a person with a larger hand and wrist has an advantage in armwrestling has less to do with strength, and more to do with leverage. In armwrestling, you and your opponent lock hands and attempt to pin each other to a pad on either side of the table. To pin your ...


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You must consider all the torques in your problem. They are four in this case : $T_f$, torque created by the force of 250 N ; $T_{P_1}$, torque created by the mass of the board on the left of pivot ; $T_{P_2}$, torque created by the mass of the board on the right of pivot ; $T_{B}$, torque created by the mass of the bucket. For compute $T_{P_1}$ and ...


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Yes, you are right! Only when a force is applied purely through the center of mass it results in the body gaining a linear action with no rotational components.When any force is applied at a distance from the center of mass, it results in the body gaining the linear acceleration mentioned above plus an angular acceleration which depends on the moment arm ...


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That's actually a really interesting question. Stating it in a slighly different way: If there is no friction, would the sphere on the inclined plane experience a torque? And the answer is - no it would not. The reason for this is that the external force on the sphere (the normal force of the plane) acts through the center of the sphere. In other words ...


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$ \newcommand{\r}{\mathbf{r}} \newcommand{\F}{\mathbf{F}} \newcommand{\g}{\mathbf{g}} \newcommand{\t}{\boldsymbol\tau} $Let me start by defining $\g(\r)$ to be a position dependent force per unit mass. Then the force per unit volume $d\F$ is given by $d\F(\r) = \g(\r) \rho (\r) dV$. The torque per unit volume $d\t$ is given by $d\t = \r \times d\F$. The ...


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The question is not detailed enough, so I am going to make some assumptions: 1)the door is vertical and on hinges; 2)there is some force (wind?) pushing on the surface of the door, creating a torque that tends to close the door; 3)a small mass is placed on the floor to try to prevent the door from closing, but it fails; 4)the small mass is replaced ...


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I'll write my comments here as a full answer, as suggested by Floris. I won't use the moment of inertia tensor: it's simpler from pure angular momentum of each point particle. We know that $$\vec{L} = (\vec{r} \times \dot{\vec{r}})\,m .$$ So, for a point particle, $$d\vec{L} = (\vec{r} \times \dot{\vec{r}})\, dm .$$ Noting that $\rho = \frac{dm}{dV}$, ...



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