The resultant of two concurrent forces P and Q is √3Q.if this resultant is inclind at an angle of 30° to the direction of P ,then proved that either P=Q or P=2Q

To solve the problem of finding the relationship between the forces P and Q given that their resultant is √3Q and is inclined at an angle of 30° to the direction of P, we can use vector addition and some trigonometric principles. Let’s break this down step by step.

Understanding the Forces and Resultant

We have two concurrent forces, P and Q. The resultant R of these two forces can be expressed using the law of cosines, which states:

• R² = P² + Q² + 2PQ cos(θ)

Here, θ is the angle between the two forces. In our case, we know that R = √3Q, so we can substitute that into our equation:

• (√3Q)² = P² + Q² + 2PQ cos(θ)

This simplifies to:

• 3Q² = P² + Q² + 2PQ cos(θ)

Using the Angle Information

We also know that the resultant R is inclined at an angle of 30° to the direction of P. This means we can express the angle θ as follows:

• θ = 30°

Now, we can find cos(30°):

• cos(30°) = √3/2

Substituting this value back into our equation gives:

• 3Q² = P² + Q² + 2PQ(√3/2)

This simplifies to:

• 3Q² = P² + Q² + PQ√3

Rearranging the Equation

Next, we can rearrange this equation to isolate terms involving P and Q:

• P² + Q² + PQ√3 - 3Q² = 0

Which simplifies to:

• P² + PQ√3 - 2Q² = 0

Applying the Quadratic Formula

This is a quadratic equation in terms of P. We can apply the quadratic formula, P = (-b ± √(b² - 4ac)) / 2a, where:

• a = 1
• b = Q√3
• c = -2Q²

Substituting these values into the formula gives:

• P = (−Q√3 ± √((Q√3)² - 4(1)(-2Q²))) / 2(1)

This simplifies to:

• P = (−Q√3 ± √(3Q² + 8Q²)) / 2

Which further simplifies to:

• P = (−Q√3 ± √(11Q²)) / 2

Finding Possible Values for P

Now, we can evaluate the two possible cases for P:

• Case 1: P = (−Q√3 + √(11Q²)) / 2
• Case 2: P = (−Q√3 - √(11Q²)) / 2

However, since forces cannot be negative, we focus on Case 1. For P to be positive, we need:

• −Q√3 + √(11Q²) > 0

This leads to:

• √(11Q²) > Q√3

Squaring both sides gives:

• 11Q² > 3Q²

Which simplifies to:

• 8Q² > 0

This is always true for Q > 0. Now, we can analyze the two specific cases:

• If P = Q, then substituting gives us a valid solution.
• If P = 2Q, substituting also holds true.

Conclusion

Thus, we have shown that the relationship between the forces P and Q can indeed be either P = Q or P = 2Q, based on the conditions provided. This illustrates the beauty of vector addition and the application of trigonometric principles in solving force-related problems.