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-2

So basically bhp and torque are irrelevant to the casual user and just a way for petrol heads to seem as if they know what they are talking about.


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Your maths is almost correct. I think you should treat the lever as two separate rods which each have a pivot at one end. If the mass per unit length is $\sigma = M/L$, then the following applies to the two rods of length $r_1$ and $r_2$ (with $r_2=L-r_1$), where I split the rod up into infinitesimal chunks of length $dr$, mass $dm=\sigma dr$ and moment of ...


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Why does torque, which is an analogy of force have the same units as energy, but force does not? The magnitude of (net) force is, in a sense, work done per unit displacement thus we can think of it as, e.g., Joules per meter. In a similar sense, the magnitude of (net) torque is work done per unit angular displacement, e.g., Joules per radian. But ...


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This is a side-effect of treating angles as dimension-less. For translational systems, we have \begin{align*} [\text{linear momentum}] &= [\text{action}][\text{length}]^{-1} \\ [\text{force}] &= [\text{linear momentum}][\text{time}]^{-1} \\&= [\text{energy}][\text{length}]^{-1} \end{align*} Correspondingly, for rotational systems, we have ...


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Torque is a cross product, and work is a dot product. So one big difference is that torque is a vector and work is a scalar. Another way to think about it is that work is a force being applied over a length interval, where only the force applied in the direction parallel to the displacement counts toward the work performed. On the other hand, torque is ...


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I'm not sure what you're referring to by 'straight' acceleration. If that angle is the angle at which a force, $F$, is being applied, then the horizontal force is: $$ F_x=F\cdot cos(\theta)\\ F_y=F\cdot sin(\theta) $$ These will get the forces along the X and Y axis respectively. Assuming width, $w$, length, $l$, height, $h$, mass, $m$ and $\text{'some ...


2

Torque is defined as $\vec \tau = \vec r \times \vec F$, where $\vec r$ is the displacement vector from the origin to the point at which the force is applied. This means that torque depends very much on the choice of origin. Then again, the choice of origin also affects the inertia tensor. So long as you get all of the physics correct, you can choose any ...


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Torque is often given as the simple equation: $$ \tau = {rFsin\theta} $$ It is the net perpendicular force which is found by finding the $sin\theta$ of the force applied where $\theta$ is the angle at which it is applied. Since torque is dependent on angular momentum and is also defined as $\tau = \frac{\partial L}{\partial t}$ The choice of origin ...


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Your question asks why there is a rotatory force that turns the coin to the right. The gravitational force is perpendicular to surface and so the vector cross-product of the gravitation force and the angular momentum (which is not exactly parallel to the surface) is perpendicular to both those vectors. That doesn't actually help (for me anyway). However, if ...


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We can start with the definition of angular momentum $\vec{L} = \vec{r} \times \vec{p}$. Differentiate both sides with respect to $t$ to get \begin{equation} \frac{d\vec{L}}{dt} = \vec{v} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt} \end{equation} The first term on the right hand side is zero because $\vec{v}$ and $\vec{p}$ are parallel to each other. ...


2

Any force not applied through the center of mass of an object will impart both linear motion and torque. You can consider the force applied at the center of mass for the computation of the linear motion, and you can then consider the torque as being the moment applied by the force at the center of mass. So for both linear and rotational motion, you end up ...


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You can sum them, without taking their respective positions into account. For example: Lets say you have a 2D body, with two forces applied to them, forming a couple. When you reference point is exactly between the points where the forces are applied, you experience $ \tau = Fa $ with $F$ as one force and $a$ as the distance between the applied forces. ...


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Force $F_E$ is downward to prevent the hand from rotating. It is due to the balance of moments. Take $F_E$ away, and the arm is going to accelerate clockwise.


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There are two force balancing considerations here. On the one hand the torque about the joint must be zero, otherwise the arm would have some angular acceleration. Treating positive forces as directed in the directions they are drawn, this tells us $$ r_1 F_\mathrm{B} = r_2 w_a + r_3 w_b. $$ Given all the distances and weights, you can solve for the tension ...


1

I don't see how the inability to expand gives rise to this force. I would explain the forces like this: In order for the situation to be stable, the total torque does not only have to be zero in the joint, but also at the point where the biceps makes contact with the bone, i.e. $$ r_1 F_E = (r_2-r_1) w_a+ (r_3-r_1) w_b\\ \Rightarrow F_E = ...


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A quick diagram: Key dimension here is the distance $d$. The weight of the robot $W = M \cdot g$ is carried equally by both legs, so we have a force $F$ along the lines $AC$ such that $$F = \frac{W}{2 \cos\alpha}$$ This force results in a torque at point $B$ because $B$ is not on the line $AC$ - it is displaced by distance $d$, given by $$d = Y ...


2

The weight vector, $W=mg$, can be split up into two components, $mg\sin\theta$ and $mg\cos\theta$, where $\theta = 30$ in your case. The $mg\sin\theta$ is the force vector which would generate a torque, $\tau = \bf{L}\times{}mg\sin\theta$.


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Torque by being the (external) product of the force applied by the radius of application represents an axial vector (or rotational vector) $$\vec{T} = \vec{F} \times \vec{r}$$ One way to represent such vector (related to the definition above) is a by a vector which is perpendicular to the plane generated by both $\vec{F}$ and $\vec{r}$ and which has a ...


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Force represents an energy transfered (and applied in general in a straight line). On the other hand Torque represents a force acting on a point on a straight line but the effects are applied elsewhere (in rotating the body). As such there is (at least) this subtle difference. The force is applied to a point of the body which then (by internal constraints ...


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I would try to explain very briefly. When you apply a force onto a body, it causes the body to move in the line of application of force. This introduces translational motion into the body. When you apply a torque onto a body, it causes the body to rotate about a point. This introduces rotational motion into the body. Torque in rotation is analogous to ...


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I suppose you could try to think of the problem in another way. What would happen to a cylinder (or sphere) if you put it on a frictionless inclined plane? Would it still roll or just slide? The imbalance in forces acting on the cylinder at different points, with respect to its center of mass, are what lead to the rotation. Gravity acts on the center of ...


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If motor A does the same job as motor B, but with a 10x greater load, and the mechanical advantage (gearing etc) is the same, then I would expect that the torque that A supplies is ten times greater as well. But that is not quite how you phrased the question. It necessarily follows that a higher HP motor can supply greater torque - at least, with the right ...



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