Problem 23 - Fundamentals of Physics

Question

Force \(\vec{F}=(2.0 \mathrm{~N}) \mathrm{i}-(3.0 \mathrm{~N}) \mathrm{k}\) acts on a pebble with position vector \(\vec{r}=(0.50 \mathrm{~m}) \mathrm{j}-(2.0 \mathrm{~m}) \hat{\mathrm{k}}\) relative to the origin. In unitvector notation, what is the resulting torque on the pebble about (a) the origin and (b) the point \((2.0 \mathrm{~m}, 0,-3.0 \mathrm{~m}) ?\)

Step-by-Step Solution

Answer

(a) Torque is \(-1.5 \mathrm{~i} + 4 \mathrm{~j} - 1 \mathrm{~k}\). (b) Torque is \(-1.5 \mathrm{~i} - 8 \mathrm{~j} - 1 \mathrm{~k}\).

Step 1: Understanding Torque Equation

The torque \( \vec{\tau} \) is given by the cross product \( \vec{\tau} = \vec{r} \times \vec{F} \), where \( \vec{r} \) is the position vector, and \( \vec{F} \) is the force vector. We need to compute this cross product to find the torque.

Step 2: Write Down Vectors for Part (a)

For the origin, the position vector is given by \( \vec{r} = 0.0 \mathrm{~i} + 0.50 \mathrm{~j} - 2.0 \mathrm{~k} \) and the force vector is \( \vec{F} = 2.0 \mathrm{~i} + 0.0 \mathrm{~j} - 3.0 \mathrm{~k} \).

Step 3: Calculating Cross Product (Part a)

Using determinant form for cross product:\[ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 0.5 & -2 \ 2 & 0 & -3 \end{vmatrix} \]This yields:\[ \vec{\tau} = (0.5 \cdot -3 - (-2) \cdot 0) \hat{i} - (0 \cdot -3 - (-2) \cdot 2) \hat{j} + (0 \cdot 0 - 0.5 \cdot 2) \hat{k} \]So, \[ \vec{\tau} = -1.5 \hat{i} + 4 \hat{j} - 1 \hat{k} \].

Step 4: Result for Part (a)

The resulting torque about the origin is \( \vec{\tau} = -1.5 \mathrm{~i} + 4 \mathrm{~j} - 1 \mathrm{~k} \).

Step 5: Compute New Position Vector for Part (b)

Find the new position vector relative to point \((2.0 \mathrm{~m}, 0, -3.0 \mathrm{~m})\):\[ \vec{r}' = (0 - 2.0) \mathrm{~i} + (0.5 - 0) \mathrm{~j} + (-2.0 - (-3.0)) \mathrm{~k} = -2.0 \mathrm{~i} + 0.5 \mathrm{~j} + 1.0 \mathrm{~k} \].

Step 6: Calculating Cross Product (Part b)

Using the new position vector and the same force vector:\[ \vec{\tau}' = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 0.5 & 1 \ 2 & 0 & -3 \end{vmatrix} \]This yields:\[ \vec{\tau}' = (0.5 \cdot -3 - 1 \cdot 0) \hat{i} - (-2 \cdot -3 - 1 \cdot 2) \hat{j} + (-2 \cdot 0 - 0.5 \cdot 2) \hat{k} \]So, \[ \vec{\tau}' = -1.5 \hat{i} - 8 \hat{j} - 1 \hat{k} \].

Step 7: Result for Part (b)

The resulting torque about point \((2.0 \mathrm{~m}, 0,-3.0 \mathrm{~m})\) is \( \vec{\tau}' = -1.5 \mathrm{~i} - 8 \mathrm{~j} - 1 \mathrm{~k} \).

Key Concepts

Cross Product

The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors in three-dimensional space. It is essential when computing the torque, which describes the rotational effect of a force.
To find the torque vector, we use the cross product of the position vector and the force vector. This operation is noted by the symbol \( \times \). This is how it works: given two vectors \( \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} \) and \( \vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k} \), their cross product is determined as follows:

• \( \vec{A} \times \vec{B} = (A_y B_z - A_z B_y) \hat{i} - (A_x B_z - A_z B_x) \hat{j} + (A_x B_y - A_y B_x) \hat{k} \)

By using a determinant, it's even easier to organize the calculation. It looks like this for our example vectors:

• \[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \end{vmatrix} \]

The cross product gives a new vector oriented perpendicular to both of the initial vectors, crucial in finding the magnitude and direction of torque.

Position Vector

A position vector is a vector that extends from a reference point (often the origin) to a specific point where a force is applied. In the context of torque calculations, it helps in determining the line of action of the force concerning the reference point.
For our problem, position vector \( \vec{r} \) is given as \( 0.0 \hat{i} + 0.50 \hat{j} - 2.0 \hat{k} \) for the origin. This vector tells us that the point of application of the force is 0.5 meters along the positive y-axis and 2.0 meters along the negative z-axis.
If the reference point changes, as in part (b) of the exercise, the position vector is recalculated to reflect the new relative distances. For example, if we shift the reference to (2.0 m, 0, -3.0 m), the new position vector \( \vec{r}' \) becomes \( -2.0 \hat{i} + 0.5 \hat{j} + 1.0 \hat{k} \). This new vector now relates the force's point of action concerning the new point of reference. Understanding the position vector is vital because the torque substantially depends on this vector's magnitude and direction.

Force Vector

In physics, a force vector represents both the magnitude and the direction of a force acting on an object. For calculating torque, understanding the force vector is vital, as it is the force that contributes to the rotational effect.
The force vector in this scenario is \( \vec{F} = 2.0 \hat{i} + 0.0 \hat{j} - 3.0 \hat{k} \). This specifies that the force has a component of 2.0 Newtons along the x-axis and -3.0 Newtons along the z-axis, with no force component in the y-direction.
Significantly, torque not only depends on the magnitude of this force but also on its direction. Therefore, a precise representation of the force vector is crucial to compute the cross product with the position vector accurately.
Being knowledgeable about both the magnitude and direction of the force helps determine how much torque is generated to cause rotational motion around a point. In various physics applications, analyzing the force vector helps us understand how objects will rotate in response to applied forces.