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.

#### Solution:

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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.

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\{2i+1.5j+0k\right\}

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

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:

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

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

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

(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(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:

#### Final Answers:

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