FM-m2-circular motion

A particle, P. of mass 3kg is attached to two light, inextensible strings. One string is attached at its other end to a point, A. The other string has its other end attached to a point, B. A is 4m above B. The particle makes horizontal circles such that angle PAB is 30° and angle PBA is 60° Given that the speed of the particle is3.6ms-l, find the tension in each string.

A particle, P. of mass 3kg is attached to two light, inextensible strings. One string is attached at its other end to a point, A. The other string has its other end attached to a point, B. A is 4m above B. The particle makes horizontal circles such that angle PAB is 30° and angle PBA is 60° Given that the speed of the particle is3.6ms-l, find the tension in each string.

To find the tension in each string, we can break the forces acting on the particle P into horizontal and vertical components and then solve for the tensions in the strings.

First, let's consider the vertical forces:

Tension in the string attached to point A (T1).
Tension in the string attached to point B (T2).
The gravitational force acting on the particle P (mg), where g is the acceleration due to gravity (approximately 9.81 m/s²).
Since the particle is moving in a horizontal circle, its vertical acceleration is zero. This means that the net vertical force acting on the particle is zero. So, we can write the following equation for the vertical forces:

T1 * cos(30°) - T2 * cos(60°) - mg = 0

Now, let's consider the horizontal forces:

Tension in the string attached to point A (T1).
Tension in the string attached to point B (T2).
The centripetal force required to keep the particle moving in a circle (F_c), which is given by F_c = (m * v²) / r, where m is the mass of the particle, v is its speed, and r is the radius of the circle.
The radius of the circle can be found by considering the triangle PAB. Using the sine rule:

sin(30°) / AB = sin(60°) / AP

sin(30°) / AB = sqrt(3) / 2 / AP

AP = AB * (2 / sqrt(3))

Now, the radius r is half of AP:

r = AB / (sqrt(3))

Now, we can write the equation for the horizontal forces:

T1 * sin(30°) + T2 * sin(60°) = F_c

Now, we can substitute the expressions for F_c and r:

T1 * sin(30°) + T2 * sin(60°) = (m * v²) / (AB / (sqrt(3)))

Now, plug in the values:

T1 * (1/2) + T2 * (sqrt(3)/2) = (3 kg * (3.6 m/s)²) / (4 m / sqrt(3))

Now, you have two equations:

T1 * cos(30°) - T2 * cos(60°) - mg = 0
T1 * (1/2) + T2 * (sqrt(3)/2) = (3 kg * (3.6 m/s)²) / (4 m / sqrt(3))
Solve these two equations simultaneously to find T1 and T2, the tensions in the strings

To find the tension in each string, we can break the forces acting on the particle P into horizontal and vertical components and then solve for the tensions in the strings.

First, let's consider the vertical forces:

Tension in the string attached to point A (T1).
Tension in the string attached to point B (T2).
The gravitational force acting on the particle P (mg), where g is the acceleration due to gravity (approximately 9.81 m/s²).
Since the particle is moving in a horizontal circle, its vertical acceleration is zero. This means that the net vertical force acting on the particle is zero. So, we can write the following equation for the vertical forces:

T1 * cos(30°) - T2 * cos(60°) - mg = 0

Now, let's consider the horizontal forces:

Tension in the string attached to point A (T1).
Tension in the string attached to point B (T2).
The centripetal force required to keep the particle moving in a circle (F_c), which is given by F_c = (m * v²) / r, where m is the mass of the particle, v is its speed, and r is the radius of the circle.
The radius of the circle can be found by considering the triangle PAB. Using the sine rule:

sin(30°) / AB = sin(60°) / AP

sin(30°) / AB = sqrt(3) / 2 / AP

AP = AB * (2 / sqrt(3))

Now, the radius r is half of AP:

r = AB / (sqrt(3))

Now, we can write the equation for the horizontal forces:

T1 * sin(30°) + T2 * sin(60°) = F_c

Now, we can substitute the expressions for F_c and r:

T1 * sin(30°) + T2 * sin(60°) = (m * v²) / (AB / (sqrt(3)))

Now, plug in the values:

T1 * (1/2) + T2 * (sqrt(3)/2) = (3 kg * (3.6 m/s)²) / (4 m / sqrt(3))

Now, you have two equations:

T1 * cos(30°) - T2 * cos(60°) - mg = 0
T1 * (1/2) + T2 * (sqrt(3)/2) = (3 kg * (3.6 m/s)²) / (4 m / sqrt(3))
Solve these two equations simultaneously to find T1 and T2, the tensions in the strings

You know that it is forbidden to show the whole solution process? it is just allowed to give clues, nothing else.

In addition, its wrong and it looks like it was done by chatgpt.

And too complicated I think. There is a right triangle with three angles and one side known (the third angle is easy to find out), if I am not mistaken. Should be the way easier to solve.