A couple is of moment G and the force forming the couple is p. If p is turned through a right angle

Q: Why is the moment a scalar? I thought moments were vectors.

In this problem, we are dealing with the magnitude of the moment of a couple. Since a couple's moment is the same about any point in the plane, we often focus on its magnitude $G = |\vec{r} \times \vec{p}| = rp \sin \theta$. This scalar approach simplifies the algebra significantly.

Q: Why does the angle change to $\theta + 90^\circ$?

Imagine the arm vector $\vec{r}$ is fixed. The force vector $\vec{p}$ makes an angle $\theta$ with it. When you rotate the force vector by $90^\circ$, you are effectively adding $90^\circ$ to the angle between the two vectors. It's a simple geometric rotation.

Q: Can I use vector notation $\vec{G} = \vec{r} \times \vec{p}$?

Yes, you absolutely can! However, since the problem asks for the relationship between the magnitudes $G, H,$ and $M$, working with the scalar magnitudes $rp \sin \theta$ is much more direct and less prone to calculation errors.

Question

A couple is of moment $\vec{G}$ and the force forming the couple is $\vec{p}$. If $\vec{p}$ is turned through a right angle the moment of the couple thus formed is $\vec{H}$. If instead, the force $\vec{p}$ are turned through an angle $\alpha$, then the moment of couple becomes

Options:

(A) $H \sin \alpha - G \cos \alpha$

(B) $G \sin \alpha - H \cos \alpha$

(C) $H \sin \alpha + G \cos \alpha$

(D) $G \sin \alpha + H \cos \alpha$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a rigid body, a silent, unyielding object waiting for you to set it into motion. You apply a couple—two equal and opposite forces that create a pure rotational effect. This is the essence of rotational mechanics.

Let us define our setup. We have an arm of the couple, represented by a vector $\vec{r}$ with length $r$, and a force $\vec{p}$ with magnitude $p$. The angle between them is $	heta$.

The moment of this couple, which we call $G$, is the magnitude of the cross product of these two vectors. Mathematically, this is:

$$G = |\vec{r} 	imes \vec{p}| = rp \sin 	heta$$

This is our foundational truth, the starting point of our journey.

### The $90^\circ$ Shift

Now, let us introduce a change. The problem asks us to rotate the force $\vec{p}$ by a right angle, $90^\circ$. Visualize the force vector swinging around the pivot point.

The new angle between the arm vector $\vec{r}$ and the force $\vec{p}$ is now $	heta + 90^\circ$. The new moment, which we call $H$, is given by:

$$H = rp \sin(	heta + 90^\circ)$$

Here is where the elegance of trigonometry comes into play. We know the identity $\sin(90^\circ + 	heta) = \cos 	heta$. Thus, our expression for $H$ simplifies beautifully to:

$$H = rp \cos 	heta$$

This is a crucial puzzle piece. We have now defined both $G$ and $H$ in terms of $r$, $p$, and $	heta$.

### The General Rotation

Finally, we consider the general case. Instead of a $90^\circ$ rotation, we rotate the force $\vec{p}$ by an arbitrary angle $\alpha$. The new angle between the arm vector $\vec{r}$ and the force $\vec{p}$ becomes $	heta + \alpha$.

Let the new moment be $M$. Following our established logic:

$$M = rp \sin(	heta + \alpha)$$

Now, we must expand this. Using the compound angle formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$, we can write:

$$M = rp (\sin 	heta \cos \alpha + \cos 	heta \sin \alpha)$$

Distributing the $rp$ term, we get:

$$M = (rp \sin 	heta) \cos \alpha + (rp \cos 	heta) \sin \alpha$$

Look closely at this expression. Do you see the familiar terms? $rp \sin 	heta$ is our initial moment $G$, and $rp \cos 	heta$ is the moment $H$ we derived earlier.

Substituting these back, we arrive at the final, elegant result:

$$\mathbf{M = G \cos \alpha + H \sin \alpha}$$

This is the beauty of physics—taking a complex rotation and reducing it to a simple, harmonious combination of known quantities. You have successfully navigated the geometry of the couple.

Animated Solution for Mathematics - Vector Algebra: A couple is of moment $G$ and the force forming the couple is $p$ . If $p$ is turned through a right angle the moment of the couple thus formed is $H$ . If instead, the force $p$ are turned through an angle $α$ , then the moment of couple becomes

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Visualized Solution

Defining the Initial Couple

Initial Moment $G$

Rotation by $9 0^{\circ}$

Moment $H$ Setup

Simplifying Moment $H$

Rotation by Angle $α$

New Moment $M$

Expanding the Sine Term

Distributing $r p$

Final Substitution

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The $9 0^{\circ}$ Shift

The General Rotation

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Visualized Solution

Defining the Initial Couple

Initial Moment .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

Rotation by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 90∘

Moment .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H Setup

Simplifying Moment .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

Rotation by Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

New Moment .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M