Q24P In unit-vector notation, what is... [FREE SOLUTION]

91影视

Chapter 11: Q24P (page 322)

In unit-vector notation, what is the torque about the origin on a jar of jalapeno peppers located at coordinates (3.0m,-2.0m,4.0m) due to (a) force ${\overset{鈬赌}{F}}_{1} = (3.0 N) \hat{i} - (4.0 N) \hat{j} + (5.0 N) \hat{k}$ ,(b) force ${\overset{鈬赌}{F}}_{2} = (- 3.0 N) \hat{i} - (4.0 N) \hat{j} + (5.0 N) \hat{k}$ , (c) the vector sum of ${\overset{鈬赌}{F}}_{1} a n d {\overset{鈬赌}{F}}_{2}$ ? (d) Repeat part (c) for the torque about the point with coordinates (3.0m, 2.0m, 4.0m).

Short Answer

Expert verified

Torque on the jar due to F₁ is $\{(6) \hat{i} + (- 3) \hat{j} + (- 6) \hat{k}\} N m$
Torque on the jar due to F₂ is $\{(26) \hat{i} + (3) \hat{j} + (- 18) \hat{k}\} N m$
Torque on the jar due to F₁ + F₂ is $\{(32) \hat{i} + (- 24) \hat{k}\} N m$
Torque about the given point is 0Nm

Step by step solution

Identification of given data

Location of jar (3.0m, -2.0m, 4.0m)

${\overset{鈬赌}{F}}_{1} = (3 \hat{i} - 4 \hat{j} + 5 \hat{k}) N {\overset{鈬赌}{F}}_{2} = (3 \hat{i} - 4 \hat{j} - 5 \hat{k}) N$

To understand the concept of torque

Using the concept of torque, the unknown torque value is calculated. As per the concept, the torque acting on a body is due to the tangential force acting on a body along a radial path of the object in a circular motion. Thus, the cross-vector of the force and radial vector of the object will give the torque value.

Formulae:

The position vector in a 3-D diagram, $\overset{鈬赌}{r} = x \hat{i} + y \hat{j} + z \hat{k}$

The force vector in 3-D, $\overset{鈬赌}{F} = F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k}$

The torque acting on the body due to the tangential force,

$\begin{array}{rcl} \overset{鈬赌}{蟿} & = & \overset{鈬赌}{r} 脳 \overset{鈬赌}{F} \\ = & |\begin{matrix} i & j & k \\ x & y & z \\ F_{x} & F_{y} & F_{z} \end{matrix}| \\ = & (y F_{z} - z F_{y}) \hat{i} + (z F_{x} - x F_{z}) \hat{j} + (x F_{y} - y F_{x}) \hat{k} \end{array}$

(a) Determining the torque about origin due to F⇀1

$\begin{array}{rcl} \overset{鈬赌}{蟿} & = & \overset{鈬赌}{r} 脳 \overset{鈬赌}{F} \\ \overset{鈬赌}{蟿} & = & |\begin{matrix} i & j & k \\ 3 & - 2 & 4 \\ 3 & - 4 & 5 \end{matrix}| \\ = & (- 10 + 16) N m \hat{i} + (12 - 15) N m \hat{j} + (- 12 + 6) N m \hat{k} \\ = & \{(6) \hat{i} + (- 3) \hat{j} + (- 6)\} \hat{k} N m \end{array}$

(b) Determining the torque about origin due to F⇀2

$\begin{array}{rcl} \overset{鈬赌}{蟿} & = & \overset{鈬赌}{r} 脳 \overset{鈬赌}{F} \\ \overset{鈬赌}{蟿} & = & |\begin{matrix} i & j & k \\ 3 & - 2 & 4 \\ - 3 & - 4 & - 5 \end{matrix}| \\ = & (10 + 16) N m \hat{i} + (- 12 + 15) N m \hat{j} + (- 12 - 6) N m \hat{k} \\ = & \{(26) \hat{i} + (3) \hat{j} + (18) \hat{k}\} N m \end{array}$

(c) Determining the torque about origin due to F⇀1+F⇀2

$\begin{array}{rcl} {\overset{鈬赌}{F}}_{1} + {\overset{鈬赌}{F}}_{2} & = & - 8 \hat{j} N \\ \overset{鈬赌}{蟿} & = & \overset{鈬赌}{r} 脳 ({\overset{鈬赌}{F}}_{1} + {\overset{鈬赌}{F}}_{2}) \\ = & |\begin{matrix} i & j & k \\ 3 & - 2 & 4 \\ 0 & - 8 & 0 \end{matrix}| \\ = & (32) \hat{i} + (0) \hat{j} + (- 24 + 0) \hat{k} \\ = & \{(32) \hat{i} + (- 24) \hat{k}\} N m \end{array}$

(d) Determining the torque about the point with coordinates (3.0m, 2.0m, 4.0m) due to F⇀1+F⇀2

So, $\overset{鈬赌}{r} = \overset{鈬赌}{r} - \overset{鈬赌}{r_{0}}$

$\overset{鈬赌}{r}' = 0 \hat{i} - 4 \hat{j} m + 0 \hat{k}$

${\overset{鈬赌}{F}}_{1} + {\overset{鈬赌}{F}}_{2} = - 8 \hat{j} N$

Torque,

$\begin{array}{rcl} \overset{鈬赌}{蟿} & = & \overset{鈬赌}{r} 脳 ({\overset{鈬赌}{F}}_{1} + {\overset{鈬赌}{F}}_{2}) \end{array}$

$\begin{array}{rcl} \overset{鈬赌}{蟿} & = & \overset{鈬赌}{r} 脳 ({\overset{鈬赌}{F}}_{1} + {\overset{鈬赌}{F}}_{2}) \\ = & |\begin{matrix} i & j & k \\ 0 & - 4 & 0 \\ 0 & - 8 & 0 \end{matrix}| \\ = & (0) \hat{i} N m + (0) \hat{j} N m + (0) \hat{k} N m \\ = & 0 N m \end{array}$