Question

One force acting on a machine part is F⃗ =(−5.00N)i^+(4.00N)j^. The vector from the origin to the point where the force is applied is r⃗ =(−0.450m)i^+(0.150m)j^.A)Use the right-hand rule to determine the direction of the torque.B)Calculate the torque produced by this force.

Answer 1

The torque's direction is determined by the right-hand rule and results in a torque vector for these vectors that points negatively along the z-axis. Using the cross product formula, the calculated torque is -1.05 N·m.

Step-by-step explanation:

To answer the questions requested by the student, we need to understand torque in a physics context. Torque, often represented by the Greek letter tau (T), is a measure of the turning force on an object. It is calculated as the cross product of the position vector (r) and the force vector (F), given by T = r x F.

A) To determine the direction of the torque using the right-hand rule, point your fingers in the direction of the position vector r, and then curl them in the direction of the force vector F. Your thumb will then point in the direction of the torque vector.

B) To calculate the torque, you take the cross product of r and F. In two dimensions with vectors given by components along the x and y axes, this simplifies to T = rxFy - ryFx. Substituting in the given values, T = (-0.450 m)(4.00 N) - (0.150 m)(-5.00 N) = -1.80 N·m + 0.75 N·m = -1.05 N·m. The negative sign indicates that the torque vector points in the negative direction of the axis perpendicular to the plane of r and F, which would be the z-axis in this case.

Answer 2

The torque magnitude is approximately
`\(2.99 \, \text{Nm}\)` and its direction is into the plane of the page.

Given:

`\[ \mathbf{F} = (-5.00 \, \text{N}) \mathbf{i} + (4.00 \, \text{N}) \mathbf{j} \]`

`\[ \mathbf{r} = (-0.450 \, \text{m}) \mathbf{i} + (0.150 \, \text{m}) \mathbf{j} \]`

1. **Direction of Torque:**

Using the right-hand rule, the thumb points into the plane of the page. Therefore, the direction of the torque
`(\(\mathbf{\tau}\)) is **(A)` Into the plane of the page.**

2. **Calculate the Torque:**

First, let's find the angle (\(\theta\)) between the force and position vectors using the dot product:

`\[ \mathbf{r} \cdot \mathbf{F} = (-0.450 \, \text{m})(-5.00 \, \text{N}) + (0.150 \, \text{m})(4.00 \, \text{N}) \] \[ \mathbf{r} \cdot \mathbf{F} = 2.25 \, \text{Nm} + 0.60 \, \text{Nm} = 2.85 \, \text{Nm} \]`

Next, find the magnitudes of the position vector
`(\(|\mathbf{r}|\))` and the force vector
`(\(|\mathbf{F}|\))`:

`\[ |\mathbf{r}| = \sqrt{(-0.450 \, \text{m})^2 + (0.150 \, \text{m})^2} = 0.474 \, \text{m} \]`

`\[ |\mathbf{F}| = \sqrt{(-5.00 \, \text{N})^2 + (4.00 \, \text{N})^2} = 6.40 \, \text{N} \]`

Now, find the angle \(\theta\):

`\[ \theta = \cos^(-1)\left(\frac{\mathbf{r} \cdot \mathbf{F}}{|\mathbf{r}| \cdot |\mathbf{F}|}\right) \]`

`\[ \theta = \cos^(-1)\left(\frac{2.85 \, \text{Nm}}{(0.474 \, \text{m})(6.40 \, \text{N})}\right) \]`

`\[ \theta \approx \cos^(-1)(0.936) \approx 19.7^\circ \]`

Finally, calculate the torque using the formula:

`\[ \mathbf{\tau} = \mathbf{r} * \mathbf{F} \] \[ \tau = |\mathbf{r}| \cdot |\mathbf{F}| \cdot \sin(\theta) \] \[ \tau \approx (0.474 \, \text{m}) \cdot (6.40 \, \text{N}) \cdot \sin(19.7^\circ) \] \[ \tau \approx 2.99 \, \text{Nm} \]`