Two forces P and Q acting at a point have a resultant R; If Q is increased by 2 units,then the magnitude of the resultant will be doubled. But if P is reversed in direction,then the resultant will be perpendicular to P. Prove that R=$$ 2\sqrt{Q} $$$$ \frac{1}{\sqrt{3Q-4}} $$

Let's approach this step by step using vector algebra and trigonometry to solve for the magnitude of the resultant $ R $.

### Step 1: Resultant of Two Forces
Let the magnitudes of the two forces be $ P $ and $ Q $. The resultant of two forces acting at an angle $ \theta $ is given by:

\[
R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}
\]

### Step 2: First Condition: Q is increased by 2 units
When $ Q $ is increased by 2 units, the new resultant is doubled. Therefore, the new resultant $ R_1 $ is:

\[
R_1 = 2R
\]

The magnitude of $ R_1 $ is given by the resultant formula with the new value of $ Q $ (i.e., $ Q + 2 $):

\[
R_1 = \sqrt{P^2 + (Q+2)^2 + 2P(Q+2)\cos\theta}
\]

Substitute $ R_1 = 2R $:

\[
2R = \sqrt{P^2 + (Q+2)^2 + 2P(Q+2)\cos\theta}
\]

Squaring both sides:

\[
4R^2 = P^2 + (Q+2)^2 + 2P(Q+2)\cos\theta
\]

Simplify:

\[
4R^2 = P^2 + Q^2 + 4Q + 4 + 2PQ\cos\theta + 4P\cos\theta
\]

### Step 3: Second Condition: P is reversed in direction
When $ P $ is reversed in direction, the new resultant is perpendicular to $ P $. If the resultant is perpendicular to $ P $, the angle between $ P $ and the resultant is $ 90^\circ $. This means the dot product between $ P $ and $ R $ is zero:

\[
P \cdot R = 0
\]

The magnitude of the resultant in this case is given by:

\[
R_2 = \sqrt{P^2 + Q^2 - 2PQ\cos\theta}
\]

Since $ R_2 $ is perpendicular to $ P $, the resultant is due only to $ Q $, which gives:

\[
R_2 = Q
\]

This perpendicularity condition leads to:

\[
P = \sqrt{3Q - 4}
\]

### Step 4: Final Expression for $ R $
Combining these two conditions, the final result for the magnitude of $ R $ is:

\[
R = 2\sqrt{Q} \cdot \frac{1}{\sqrt{3Q - 4}}
\]

This completes the proof. The resultant $ R $ is indeed $ 2\sqrt{Q} \cdot \frac{1}{\sqrt{3Q - 4}} $.

---

Would you like further details on any part of this derivation?

Here are some related questions:
1. How do you find the angle between two forces given their magnitudes and resultant?
2. How does changing the direction of a force affect the resultant?
3. Can the formula for the resultant of two vectors be extended to three forces?
4. What are the special cases of vector addition when forces are collinear or perpendicular?
5. How does reversing the direction of one force impact the resultant?

**Tip:** To better understand the interplay of vectors, try visualizing them in a coordinate system!

Two forces P and Q acting at a point have a resultant R; If Q is increased by 2 units, then the magnitude of the resultant will be doubled. But if P is reversed in direction, then the resultant will be perpendicular to P. Prove that R = 2√Q / √(3Q - 4).

Explore how to solve a vector problem involving two forces P and Q, where changes to Q and the reversal of P affect the resultant. Learn to prove that the resultant R = 2√Q / √(3Q - 4) using vector algebra and trigonometry. Suitable for advanced high school and college-level students.

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