# Using a ring collar the 75-N force

Using a ring collar the 75-N force can act in the vertical plane at various angles θ. Determine the magnitude of the moment it produces about point A, plot the result of M (ordinate) versus θ (abscissa) for 0º ≤ θ ≤ 180º and specify the angles that give the maximum and minimum moment.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

To determine the moment created by the force at the ring colar, we first have to express 75 N in Cartesian vector form. This force can be broken into two components, a y-component and a z-component. Notice how it does not have a x-component because it lies on the z-y plane.

y-component = $75\cos\theta$

z-component = $75\sin\theta$

Thus, the force in carteisan vector form is:

$F=\left\{0i+75\cos\theta j+75\sin\theta k\right\}$

We can now draw a position vector from A to the ring collar as follows:

This position vector is:

$r_{AB}\,=\,\left\{(2-0)i+(1.5-0)j+(0-0)k\right\}$

$r_{AB}=\left\{2i+1.5j+0k\right\}$

A position vector, denoted $\mathbf{r}$ is a vector beginning from one point and extending to another point. It is calculated by subtracting the corresponding vector coordinates of one point from the other. If the coordinates of point A was $(x_A,y_A,z_A)$ and the coordinates of point B was$(x_B,y_B,z_B)$, then $r_{AB}\,=\,(x_B-x_A)i+(y_B-y_A)j+(z_B-z_A)k$

Taking the cross product between the position vector, $r_{AB}$ and the force, $F$ gives us the moment at A.

$M_A=r_{AB}\times F$

$M_A=\begin{bmatrix}\bold i&\bold j&\bold k\\2&1.5&0\\0&75\cos\theta&75\sin\theta\end{bmatrix}$

$M_A=112.5\sin\theta i-150\sin\theta j+150\cos\theta k$

The magnitude of this moment is:

$M_{magnitude}=\sqrt{(112.5\sin\theta)^2+(-150\sin\theta)^2+(150\cos\theta)^2}$

(Remember the trigonometric identity which states: $\sin^2\theta+\cos^2\theta=1$)

$M_{magnitude}=\sqrt{12656.25\sin^2\theta+22500}$

The magnitude is equal to the square root of the sum of the squares of the vector. If the position vector was $r\,=\,ai+bj+ck$, then the magnitude would be, $r_{magnitude}\,=\,\sqrt{(a^2)+(b^2)+(c^2)}$. In the simplest sense, you take each term of a vector, square it, add it together, and then take the square root of that value.

To find the minimum and maximum, we have to take the derivative of the function and set it to zero.

$\dfrac{\text{d}(\sqrt{12656.25\sin^2\theta+22500})}{\text{d}y}=\dfrac{6328.125\sin(2\theta)}{\sqrt{12656.25\sin^2(\theta)+22500}}$

(Set it equal to 0 to find the maximum and minimum)

$0=\dfrac{6328.125\sin(2\theta)}{\sqrt{12656.25\sin^2(\theta)+22500}}$

$\theta=0^0,90^0,180^0$

Derivative solved
Maximums and minimums solved

Substituting these values into our original equation gives us the values of the said maximum and minimum.

$M_A=\sqrt{12656.25\sin^2(90^0)+22500}=187.5N\cdot m$

$M_A=\sqrt{12656.25\sin^2(180^0)+22500}=150N\cdot m$

$M_A=\sqrt{12656.25\sin^2(0^0)+22500}=150N\cdot m$

The graph of this equation between 0º ≤ θ ≤ 180º is as follows:

When $\theta=0^0, 180^0, M_A=150N\cdot m$
When $\theta=90^0, M_A=187.5N\cdot m$