two forces are in the ratio 1:2 named p and q.if their resultant is at an angle tan inverse root 3 /2 to vector p then the angle between p and q is ?
Plz solve.
To solve this problem, we need to use vector addition and trigonometry. Let's start by assigning magnitudes to the forces and expressing them in terms of some variable.
Let the magnitude of force p be x, and the magnitude of force q be 2x (since they are in the ratio 1:2).
Now, let's find the resultant of these two forces. The resultant is obtained by adding the two vectors together. Since we are given the angle between the resultant and force p, we can use this information to determine the direction of the resultant vector.
Given that the angle between the resultant vector and force p is tan inverse(root 3/2), we can determine the direction of the resultant vector. The tangent of an angle is equal to the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle in a right triangle.
In this case, the side opposite the angle is the magnitude of force p, which is x, and the side adjacent to the angle is the magnitude of the resultant vector. Let's call the magnitude of the resultant vector R. So, we have:
tan(tan inverse(root 3/2)) = x/R
Simplifying the equation, we get:
root 3/2 = x/R
Multiplying both sides by R, we obtain:
R * (root 3/2) = x
Now, we have the magnitude of the resultant vector R in terms of x. Next, we need to find the angle between force p and force q.
To find this angle, we can use the law of cosines. In a triangle, the square of one side is equal to the sum of the squares of the other two sides, minus twice the product of the magnitudes of the two sides multiplied by the cosine of the angle between them.
In this case, we have a triangle with sides x (force p), 2x (force q), and R (resultant vector). The angle between force p and force q is the angle opposite to the resultant vector R.
Applying the law of cosines, we have:
R^2 = x^2 + (2x)^2 - 2(x)(2x) * cos(angle between p and q)
Simplifying the equation, we get:
R^2 = 5x^2 - 4x^2 * cos(angle between p and q)
Now, substituting the value of R from the previous equation (R * (root 3/2) = x), we get:
(x * (root 3/2))^2 = 5x^2 - 4x^2 * cos(angle between p and q)
Simplifying further, we have:
3x^2/4 = 5x^2 - 4x^2 * cos(angle between p and q)
Canceling out x^2, we obtain:
3/4 = 5 - 4 * cos(angle between p and q)
Rearranging the equation, we get:
4 * cos(angle between p and q) = 5 - 3/4
4 * cos(angle between p and q) = 20/4 - 3/4
4 * cos(angle between p and q) = 17/4
Dividing both sides by 4, we have:
cos(angle between p and q) = 17/16
Now, we can find the angle between force p and force q by taking the inverse cosine (cos^-1) of 17/16:
angle between p and q = cos^-1(17/16)
Calculating this inverse cosine, we get:
angle between p and q ≈ 23.22 degrees
Therefore, the angle between force p and force q is approximately 23.22 degrees.