The force 𝐅=2 𝐢-5 𝐤 acts at the point (3,-1,0) . Find the torque of 𝐅 about each of the followin

The force 𝐅=2 𝐢-5 𝐤 acts at the point (3,-1,0) . Find the...

The force $\mathbf{F}=2 \mathbf{i}-5 \mathbf{k}$ acts at the point $(3,-1,0) .$ Find the torque of $\mathbf{F}$ about each of the following lines. (a) $\quad \mathbf{r}=(2 \mathbf{i}-\mathbf{k})+(3 \mathbf{j}-4 \mathbf{k}) t$. (b) $\quad \mathbf{r}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}+(2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}) t$.

The force $\mathbf{F}=2 \mathbf{i}-5 \mathbf{k}$ acts at the point $(3,-1,0) .$ Find the torque of $\mathbf{F}$ about each of the following lines.
(a) $\quad \mathbf{r}=(2 \mathbf{i}-\mathbf{k})+(3 \mathbf{j}-4 \mathbf{k}) t$.
(b) $\quad \mathbf{r}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}+(2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}) t$.

Key Concepts

Parametric Representation of a Line

A line in three-dimensional space is typically represented in parametric form, which expresses the line as a point plus a scalar multiple of a direction vector. This form not only conveys a point on the line but also clearly indicates the direction in which the line extends. Identifying the direction vector is essential when calculating the moment about the line, as it is used in the projection process.

Cross Product

The cross product is a vector operation that combines two vectors to produce a third vector, which is perpendicular to the plane formed by the original vectors. Its magnitude is proportional to the area of the parallelogram defined by the two vectors, and it is a fundamental tool in physics for computing quantities like torque, where the direction of the resulting vector indicates the axis of rotation according to the right-hand rule.

Torque (Moment) of a Force

This concept involves the idea that a force acting at a point not located at the axis of rotation will produce a turning effect about that axis. The torque (or moment) vector is defined as the cross product of the position vector (from the point or line of reference to the point where the force is applied) and the force vector. It quantifies how effective the force is at producing rotational motion about a given point or axis.

Moment About a Line (Axis)

The moment of a force about a line (or axis) is obtained by projecting the torque vector onto the unit vector along that line. This projection gives the component of the torque that actually causes rotation about the axis. It is used to assess the effectiveness of a force in producing rotation about specific lines or axes in three-dimensional space.

Transcript

00:01 In this question, we are given a force 2 i -hat minus 5 k -hat acting at the point 3 minus 1, 0, and we're asked to find the torque due to that force about each of these two lines.

00:22 So in general, recall that the scalar torque about a line due to the force f is given by the formula n dot r cross f, where f is the vector of that force, n is a unit vector along the direction of the line, and r is a vector going from any point on the line to any point on the line of action of f.

00:51 So let's first find n and r that we can use.

01:04 Note that there are an infinite number of possible r's that we can use, and two possible n's corresponding to the positive and the negative version.

01:17 So first of all, n.

01:20 In part a, this is our line, and it has a direction vector given by 3j minus 4k, and so n is just, again, either plus or minus, but i'll just say plus that direction vector divided by its magnitude, since we need a unit vector.

01:44 Now we know the magnitude of a vector with components 3 and 4 is 5, because the square root of 3 squared plus 4 squared is 25, or is the square root of 25, or 5.

02:04 Now for r.

02:07 Note that it's not the same thing as this r, which is just a general vector on the line used to express that vector equation.

02:21 Our r is going to go from any point on the line of, the line about which we're calculating, to any point on the line of action of now a good choice of a line on the, a point on the line, is this one, given by t equals zero...

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