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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.
Reply 1
Original post by Lacey-yu
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.


what have you done so far?
Is it possible for you to attach a sketch? Would be nice to know, whether you have done any calculations or at least made thoughts so far.
(edited 9 months ago)
Is it a conical pendulum by any chance? if so, you just need to find out the radius in this pendulum. My clue to you: if the other angle is 30° and the other one 60°, what is the value of the third one in the triangle PAB? you maybe are able to use a well known law for that kind of triangle...
Reply 4
Original post by Lacey-yu
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
(edited 9 months ago)
Reply 5
Original post by frankin
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


Its worth having a read of the posting guidlines sticky at the top of the forum about not posting "solutions". Also the solution should be a couple of lines (with a diagram) if you resolve forces along each string as theyre perpendicular/indepenent and you can pretty much write the tensions down. Using the sin rule etc simply unnecessary in a right triangle and the radius looks wrong ...
(edited 9 months ago)
Original post by frankin
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.
(edited 9 months ago)
Reply 7
Original post by Kallisto
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.
Original post by mqb2766
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.
(edited 9 months ago)
Reply 9
Original post by Kallisto
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.

Chatgpt should have sketched the diagram.

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