A triangular plate is shown below. A force F = 4i - 3j is applied at point P. The torque at point P

Q: Why is the position vector $\mathbf{r}_Q$ calculated as $\mathbf{r}_P - \mathbf{r}_Q$?

The position vector must always point from the pivot (the point about which you are calculating torque) to the point where the force is applied. So, relative to Q, the position of P is $\mathbf{r}_P - \mathbf{r}_Q$.

Q: How do we know the coordinates of P are $(5, 5\sqrt{3})$?

The triangle has a base of 10 cm and base angles of $60^\circ$, making it an equilateral triangle. Dropping a perpendicular from P to the base bisects it at $x = 5$. The height is $10 \sin 60^\circ = 5\sqrt{3}$.

Q: What happens to the $\hat{i} \times \hat{i}$ and $\hat{j} \times \hat{j}$ terms in the cross product?

The cross product of any vector with itself is zero because the angle between them is $0^\circ$ and $\sin 0^\circ = 0$. Thus, $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{j} = 0$.

Q: Why does the book say the answer is (b)?

The book's answer key contains a typographical error. If you carefully follow the cross product math, the torque about O is strictly $-(15 + 20\sqrt{3})\hat{k}$, which matches option (a). Trust the math!

Q: Does the torque depend on the choice of the origin?

Yes! Torque is always defined relative to a specific pivot point. As we saw, shifting the pivot from O to Q changed the position vector, which in turn completely changed the resulting torque.

Question

A triangular plate is shown below. A force $\mathbf{F} = 4\hat{i} - 3\hat{j}$ is applied at point P. The torque at point P with respect to point O and Q are

Options:

(A) $-15 - 20\sqrt{3}, 15 - 20\sqrt{3}$

(B) $15 + 20\sqrt{3}, 15 - 20\sqrt{3}$

(C) $15 - 20\sqrt{3}, 15 + 20\sqrt{3}$

(D) $-15 + 20\sqrt{3}, 15 + 20\sqrt{3}$

Sigma Percentile · Accepted Answer

### Setting the Stage: The Geometry of the Plate

Before we can calculate any torques, we need to understand the exact geometry of the triangular plate. The problem provides us with a triangle where the base $OQ$ lies on the X-axis and has a length of $10	ext{ cm}$. The angles at vertices $O$ and $Q$ are both $60^\circ$. This tells us that the triangle is equilateral!

Let's set up our coordinate system with point $O$ at the origin $(0,0)$. Since $Q$ is $10	ext{ cm}$ away along the X-axis, its coordinates are $(10, 0)$. To find the coordinates of point $P$, we can use basic trigonometry. The x-coordinate is $10 \cos 60^\circ = 5$, and the y-coordinate is $10 \sin 60^\circ = 5\sqrt{3}$. Therefore, the position of $P$ is $(5, 5\sqrt{3})$.

### The Master Equation: Torque as a Cross Product

Torque is a measure of the rotational force applied to an object. Mathematically, the torque $\mathbf{	au}$ about a specific pivot point is defined as the cross product of the position vector $\mathbf{r}$ (from the pivot to the point of force application) and the force vector $\mathbf{F}$.

$$\mathbf{	au} = \mathbf{r} 	imes \mathbf{F}$$

In our problem, the force applied at point $P$ is given as $\mathbf{F} = 4\hat{i} - 3\hat{j}$. Our goal is to calculate the torque about two different pivot points: $O$ and $Q$.

### Calculating Torque about the Origin (Point O)

Let's start with point $O$. The position vector of $P$ relative to $O$ is simply the coordinate vector of $P$ itself:

$$\mathbf{r}_O = 5\hat{i} + 5\sqrt{3}\hat{j}$$

Now, we compute the cross product $\mathbf{	au}_O = \mathbf{r}_O 	imes \mathbf{F}$:

$$\mathbf{	au}_O = (5\hat{i} + 5\sqrt{3}\hat{j}) 	imes (4\hat{i} - 3\hat{j})$$

Remember the rules of the cross product for unit vectors: $\hat{i} 	imes \hat{i} = 0$, $\hat{j} 	imes \hat{j} = 0$, $\hat{i} 	imes \hat{j} = \hat{k}$, and $\hat{j} 	imes \hat{i} = -\hat{k}$. Expanding the terms:

$$\mathbf{	au}_O = 5(-3)(\hat{i} 	imes \hat{j}) + 5\sqrt{3}(4)(\hat{j} 	imes \hat{i})$$

$$\mathbf{	au}_O = -15\hat{k} - 20\sqrt{3}\hat{k} = -(15 + 20\sqrt{3})\hat{k}$$

### Shifting the Pivot: Torque about Point Q

Next, we calculate the torque about point $Q$. This time, the position vector must originate from $Q$ and point to $P$. We find this by subtracting the coordinates of $Q$ from $P$:

$$\mathbf{r}_Q = \mathbf{r}_P - \mathbf{r}_Q = (5\hat{i} + 5\sqrt{3}\hat{j}) - 10\hat{i} = -5\hat{i} + 5\sqrt{3}\hat{j}$$

Now, we compute the cross product $\mathbf{	au}_Q = \mathbf{r}_Q 	imes \mathbf{F}$:

$$\mathbf{	au}_Q = (-5\hat{i} + 5\sqrt{3}\hat{j}) 	imes (4\hat{i} - 3\hat{j})$$

Expanding the terms just like before:

$$\mathbf{	au}_Q = (-5)(-3)(\hat{i} 	imes \hat{j}) + 5\sqrt{3}(4)(\hat{j} 	imes \hat{i})$$

$$\mathbf{	au}_Q = 15\hat{k} - 20\sqrt{3}\hat{k} = (15 - 20\sqrt{3})\hat{k}$$

### The Final Verdict

We have found the torques about both points. The z-component of the torque about $O$ is $-15 - 20\sqrt{3}$, and the z-component of the torque about $Q$ is $15 - 20\sqrt{3}$. Looking at our options, this perfectly matches option (a).

*(Note: Some reference materials or answer keys might contain a typographical error pointing to option (b), but as we've rigorously proven with the cross product, option (a) is the mathematically correct answer!)*

Animated Solution for Physics - Rotational Motion: A triangular plate is shown below. A force $F = 4 \hat{i} - 3 \hat{j}$ is applied at point P. The torque at point P with respect to point O and Q are

Select Answer:

Visualized Solution

Analyzing the Geometry

The Torque Formula

Position Vector relative to $O$

Torque about $O$

Position Vector relative to $Q$

Torque about $Q$

Final Conclusion

The Sigma Insight: Torque and Angular Momentum

Setting the Stage

The Master Equation

Calculating Torque about the Origin (Point O)

Shifting the Pivot

The Final Verdict

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