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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 ...


1

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 ...


1

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 ...


1

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 ...


1

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 ...


-1

Actually with the lever sticking up above the balance scale shown; balance axis point; it adds gravitational weight to whichever side is initially positioned downward, and also since this side is down, it should weigh more being closer to the ground... and should tend to keep this side down... However is doesn't... If you put a weight, on some point of a ...


1

The problem with attempting to do the analysis with the forward point of contact on the box when it is sliding is that the box is accelerating. This makes a non-inertial frame and there's more moving parts. Besides the force of gravity on the center of mass, there will be fictitious forces. First, lets assume friction is zero. If so, we can calculate ...


0

Take a look on the forces diagram in the first picture. Let's ignore the force $N$ for now, it can only contribute to the tipping (due to its orientation), but since it is going to be confined to acting only on the red point as soon as the tipping occurs (if it does), it won't have any influence at all. E.g. it produces zero torque about the red point. ...


1

It's part and parcel of the definitions of angular momentum and torque. Consider a system of particles, not necessarily a rigid body. Without loss of generality, we can use an inertial frame that is instantaneously co-located and co-moving with the system center of mass. The angular momentum of the system with respect to the origin of this frame is defined ...


1

The physical explanation for why torque increases with r is that the longer the lever arm is the greater angular acceleration you can cause for a given force F. If a screw is stuck because it was screwed in too hard (ie with too much torque), you need to get a longer wrench. With the longer wrench (ie, larger r_w) you can generate greater force at the edge ...


1

Why would there be no radius in torque? A real torque is a real force that acts on a real rotating body (rigid or not) at a real radius. If you look at e.g. rotating machine parts, they all have a finite diameter. That diameter is of enormous importance for the design of a part, because together with the material constants it determines just how much torque ...


2

Can a ball stay still while laying on a inclined plane? In freshmen physics, the inclined plane and ball are perfect and the ball moves, so for your purposes, no. If either the surface or the ball have imperfections, we can tip the plane and the ball won't move until gravity exceeds the sum of the normal forces. To imagine those normal forces, we look very ...


3

General remarks. That's right. Torque is defined as $\mathbf r\times\mathbf F$ where $\mathbf r$ is the position where the force is applied, and $\mathbf F$ is the force being applied. The so-called Law of the Lever can then be derived from the following fact (which itself can be derived from Newton's Laws) about systems of particles: The net external ...


0

As you say, torque can be defined as $ r \times F $. Then the law of the lever is derived simply from the condition that at static equilibrium all forces on the lever and all torques about the fulcrum sum to zero.



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