Two forces 3 N and 2 N are at angle  such that theresultant is R. The first force is now increased to 6N and the resultant become 2R. The value of  is

The problem involves vector addition and trigonometry. Here are the steps to solve it:

Step 1: The resultant of two forces can be found using the formula: R = sqrt((F1^2) + (F2^2) + 2F1F2*cos(θ)), where F1 and F2 are the magnitudes of the forces and θ is the angle between them.

Step 2: Substitute the given values into the formula: R = sqrt((3^2) + (2^2) + 232*cos(θ)) = sqrt(13 + 12cos(θ)).

Step 3: The problem states that when the first force is increased to 6N, the resultant becomes 2R. Substitute these values into the formula: 2R = sqrt((6^2) + (2^2) + 262*cos(θ)) = sqrt(40 + 24cos(θ)).

Step 4: Since 2R = sqrt(40 + 24cos(θ)), we can square both sides to get rid of the square root: 4R^2 = 40 + 24cos(θ).

Step 5: Substitute R^2 from step 2 into the equation from step 4: 4*(13 + 12cos(θ)) = 40 + 24cos(θ).

Step 6: Simplify the equation to solve for cos(θ): 52 + 48cos(θ) = 40 + 24cos(θ), which simplifies to 24cos(θ) = -12.

Step 7: Solve for cos(θ): cos(θ) = -12/24 = -0.5.

Step 8: Use the inverse cosine function to find the angle θ: θ = cos^(-1)(-0.5) = 120 degrees or 2π/3 radians.

So, the value of θ is 120 degrees or 2π/3 radians.