Question

To understand the two most common procedures for finding torques when the forces and displacements are all in one plane: the moment arm method and the tangential force method. The purpose of this problem is to give you further practice finding torques in two-dimensional situations. In this case it is overkill to use the full cross product definition of the torque because the only nonzero component of the torque is the component perpendicular to the plane containing the problem. There are two common methods for finding torque in a two-dimensional problem: the tangential force methodand the moment arm method. Both of these methods will be illustrated in this problem. Throughout the problem, torques that would cause counterclockwise rotation are considered to be positive. Consider a uniform pole of length L, attached at its base (via a pivot) to a wall. The other end of the pole is attached to a cable, so that the pole makes an angle? with respect to the wall, and the cable is horizontal. The tension in the cable is T. The pole is attached to the wall at point A. (Figure 1)

a) What is Ft, the magnitude of the tangential force that acts on the pole due to the tension in the rope? (Express your answer in terms of T and ?.) When using the tangential force method, you calculate the torque using the equation ?=Ftd, where d is the distance from the pivot to the point where the force is applied. The sign of the torque can be determined by checking which direction the tangential force would tend to cause the pole to rotate (where counterclockwise rotation implies positive torque).

b) What is the magnitude of the torque ? on the pole, about point A, due to the tension in the rope? (Express your answer in terms of T, L, and ?.) Moment arm method The moment arm method involves finding the effective moment arm of the force. To do this, imagine a line parallel to the force, running through the point at which the force is applied, and extending off to infinity in either direction. You may shift the force vector anywhere you like along this line without changing the torque, provided you do not change the direction of the force vector as you shift it. It is generally most convenient to shift the force vector to a point where the displacement from it to the desired pivot point is perpendicular to its direction. This displacement is called the moment arm. For example, consider the force due to tension acting on the pole. Shift the force vector to the left, so that it acts at a point directly above the point A in the figure. The moment arm of the force is the distance between the pivot and the tail of the shifted force vector. The magnitude of the torque about the pivot is the product of the moment arm and force, and the sign of the torque is again determined by the sense of the rotation of the pole it would cause. (Figure 3)

c) Find Rm, the length of the moment arm of the force. (Express your answer in terms of L and ?.) Now consider a woman standing on the ball of her foot as shown (Figure 4) . A normal force of magnitude N acts upward on the ball of her foot. The Achilles' tendon is attached to the back of the foot. The tendon pulls on the small bone in the rear of the foot with a force F. This small bone has a length x, and the angle between this bone and the Achilles' tendon is ?. The horizontal displacement between the ball of the foot and the point P is D.

d) Suppose you were asked to find the torque about point P due to the normal force N in terms of given quantities. Which method of finding the torque would be the easiest to use? Tangential force method or moment arm method.

e) Suppose you were asked to find the torque about point P due to the force of magnitude F in the Achilles' tendon. Which of the following statements is correct?

Answer 1

Concepts and reason

The method required to solve this problem are tangential force method and moment arm method.

Initially use the trigonometric function to solve for the tangential component of the force.

Later use the torque equation used in the tangential force method to solve for torque due to tension force. Then use the trigonometric function to solve for the moment arm.

Finally identify the method that can be used for the given problem.

Fundamentals

The tangential force method involves finding the component of the applied force that is perpendicular to the displacement from the pivot point to where the force is applied. This perpendicular component of the force is called the tangential force.

The torque equation used in the tangential force method is,

$\tau = {F_{\rm{t}}}d$

Here, ${F_{\rm{t}}}$ is the tangential force, and d is the distance from the pivot point to the point where force is applied.

The moment arm method involves finding the effective moment arm of the force. The displacement is called the moment arm.

The torque equation used in the moment arm method is,

$\tau = F{R_{\rm{m}}}$

Here, $F$ is the force, and ${R_{\rm{m}}}$ is the moment arm.

The trigonometric identity is as follows:

$\cos \theta = \frac{{{\rm{adjacent}}}}{{{\rm{hypotenuse}}}}$

Here, $\theta$ is the angle, adjacent is the side adjacent to the angle $\theta$ , and hypotenuse is the longest side of the right-angled triangle.

The trigonometric identity is as follows:

$\sin \phi = \frac{{{\rm{opposite}}}}{{{\rm{hypotenuse}}}}$

Here, $\phi$ is the angle, opposite is the side opposite to the angle $\theta$ , and hypotenuse is the longest side of the right-angled triangle.

(a)

Draw the sketch of the diagram to find the tangential force ${F_{\rm{t}}}$ . The Tension in the rope is $T$ . The angle between the tension and tangential force is $\theta$ .

Use cosine function to solve for the tangential component of the Tension force $T$ .

Substitute $T$ for ${\rm{hypotenuse}}$ , and ${F_{\rm{t}}}$ for adjacent in the equation $\cos \theta = \frac{{{\rm{adjacent}}}}{{{\rm{hypotenuse}}}}$ and rearrange to solve for ${F_{\rm{t}}}$ .

$\begin{array}{c}\\\cos \theta = \frac{{{F_{\rm{t}}}}}{T}\\\\{F_{\rm{t}}} = T\cos \theta \\\end{array}$

(b)

Use the torque equation of tangential force method.

Substitute $T\cos \theta$ for ${F_{\rm{t}}}$ , and $L$ for $d$ in the equation $\tau = {F_{\rm{t}}}d$ .

$\begin{array}{c}\\\tau = \left( {T\cos \theta } \right)L\\\\ = TL\cos \theta \\\end{array}$

(c)

Refer the diagram in the question with ${R_{\rm{m}}}$ . The ${R_{\rm{m}}}$ is the adjacent to the angle $\theta$ and $L$ is the hypotenuse.

Use cosine function to solve for the ${R_{\rm{m}}}$ .

Substitute $L$ for ${\rm{hypotenuse}}$ , and ${R_{\rm{m}}}$ for adjacent in the equation $\cos \theta = \frac{{{\rm{adjacent}}}}{{{\rm{hypotenuse}}}}$ and rearrange to solve for ${R_{\rm{m}}}$ .

$\begin{array}{c}\\\cos \theta = \frac{{{R_{\rm{m}}}}}{L}\\\\{R_{\rm{m}}} = L\cos \theta \\\end{array}$

Use the torque equation of moment arm method.

Substitute $T$ for $F$ , and $L\cos \theta$ for ${R_{\rm{m}}}$ in the equation $\tau = F{R_{\rm{m}}}$ .

$\tau = TL\cos \theta$

(d)

Determine the torque at point P due to normal force.

Torque about a point is given as,

$\tau = F{R_{\rm{m}}}$

Substitute -N for F and D for ${R_{\rm{m}}}$ in the above equation.

$\tau = - ND$

Here, N is the normal force and D is the length of the moment arm of the force.

(e)

Determine the torque about point P due to the force of magnitude F by using both the methods.

Using tangential force method:

Use cosine function to solve for the tangential component of the force F.

Substitute F for ${\rm{hypotenuse}}$ , and ${F_{\rm{t}}}$ for opposite in the equation $\sin \phi = \frac{{{\rm{opposite}}}}{{{\rm{hypotenuse}}}}$ and rearrange to solve for ${F_{\rm{t}}}$ .

$\begin{array}{c}\\\sin \phi = \frac{{{F_{\rm{t}}}}}{F}\\\\{F_{\rm{t}}} = F\sin \phi \\\end{array}$

Use torque equation of tangential force method.

Substitute x for d and $F\sin \phi$ for ${F_{\rm{t}}}$ in the equation $\tau = d{F_{\rm{t}}}$ .

$\tau = xF\sin \phi$

Using moment arm method:

Draw perpendicular from the pivot point P to the line of action of force.

Use sine function to solve for the ${R_{\rm{m}}}$ .

Substitute x for hypotenuse, and ${R_{\rm{m}}}$ for opposite in the equation $\sin \phi = \frac{{{\rm{opposite}}}}{{{\rm{hypotenuse}}}}$ and rearrange to solve for ${R_{\rm{m}}}$ .

$\begin{array}{c}\\\sin \phi = \frac{{{R_{\rm{m}}}}}{x}\\\\{R_{\rm{m}}} = x\sin \phi \\\end{array}$

Use the torque equation of moment arm method.

Substitute $x\sin \phi$ for ${R_{\rm{m}}}$ in the equation $\tau = F{R_{\rm{m}}}$ .

$\tau = Fx\sin \phi$

Hence, either method can be used to find the torque.

Ans: Part a

The magnitude of tangential force that acts on the pole due to the tension in the rope is ${F_{\rm{t}}} = T\cos \theta$ .