Math Problem Statement

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}}

Solution

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

Step 1: Resultant of Two Forces

Let the magnitudes of the two forces be PP and QQ. The resultant of two forces acting at an angle θ\theta is given by:

R=P2+Q2+2PQcosθR = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}

Step 2: First Condition: Q is increased by 2 units

When QQ is increased by 2 units, the new resultant is doubled. Therefore, the new resultant R1R_1 is:

R1=2RR_1 = 2R

The magnitude of R1R_1 is given by the resultant formula with the new value of QQ (i.e., Q+2Q + 2):

R1=P2+(Q+2)2+2P(Q+2)cosθR_1 = \sqrt{P^2 + (Q+2)^2 + 2P(Q+2)\cos\theta}

Substitute R1=2RR_1 = 2R:

2R=P2+(Q+2)2+2P(Q+2)cosθ2R = \sqrt{P^2 + (Q+2)^2 + 2P(Q+2)\cos\theta}

Squaring both sides:

4R2=P2+(Q+2)2+2P(Q+2)cosθ4R^2 = P^2 + (Q+2)^2 + 2P(Q+2)\cos\theta

Simplify:

4R2=P2+Q2+4Q+4+2PQcosθ+4Pcosθ4R^2 = P^2 + Q^2 + 4Q + 4 + 2PQ\cos\theta + 4P\cos\theta

Step 3: Second Condition: P is reversed in direction

When PP is reversed in direction, the new resultant is perpendicular to PP. If the resultant is perpendicular to PP, the angle between PP and the resultant is 9090^\circ. This means the dot product between PP and RR is zero:

PR=0P \cdot R = 0

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

R2=P2+Q22PQcosθR_2 = \sqrt{P^2 + Q^2 - 2PQ\cos\theta}

Since R2R_2 is perpendicular to PP, the resultant is due only to QQ, which gives:

R2=QR_2 = Q

This perpendicularity condition leads to:

P=3Q4P = \sqrt{3Q - 4}

Step 4: Final Expression for RR

Combining these two conditions, the final result for the magnitude of RR is:

R=2Q13Q4R = 2\sqrt{Q} \cdot \frac{1}{\sqrt{3Q - 4}}

This completes the proof. The resultant RR is indeed 2Q13Q42\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!

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Math Problem Analysis

Mathematical Concepts

Vector Addition
Trigonometry
Resultant of Forces

Formulas

R = √(P^2 + Q^2 + 2PQ cosθ)
P · R = 0
R = 2√Q / √(3Q - 4)

Theorems

Pythagorean Theorem
Law of Cosines

Suitable Grade Level

Grade 11-12, College Level