# Thread: the angles in a triangle of force

1. ## the angles in a triangle of force

hello,

I'm currently studying problems involving forces in equilibrium.

In my text book, I'm examining 3 different ways of solving the same problem. In this particular question, you have to find the values of T1 and T2.

In the part I'm on now, you solve it by drawing a Triangle of Force.

I understand how to do the Vectors, but I don't know how they've worked out the angles in the triangle. Could anyone manage to explain this to me at all?

To describe it, the force diagram consists of an object of mass 12 Kg,suspended by 2 strings called T1 and T2. String T1 is sort of pointing upwards and leftwards, forming an angle of 27 degrees with the horizon at the top. String T2 is pointing upwards and rightwards, forming an angle of 38 degrees with the horizon at the top. At the bottom of the 2 strings, where they meet, hangs the object of 12gN.

IN THE TRIANGLE THAT I CAN NOW SEE UNDER MY POST, THE ANGLES IN THEIR FORCE TRIANGLE ARE:
BOTTOM LEFT - 52 degrees
RIGHT - 65 degrees
TOP - 63 degree
The vectors in the triangle of force are - VERTICAL ONE ON LEFT = 12g. ONE AT TOP = T1. ONE STARTING AT BOTTOM AND SLOPING UPWARDS AND RIGHTWARDS = T2[/U]
[IMG]file:///C:/Users/Moi/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]
[IMG]file:///C:/Users/Moi/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]
[IMG]file:///C:/Users/Moi/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]

2. Originally Posted by protractor
hello,

I'm currently studying problems involving forces in equilibrium.

In my text book, I'm examining 3 different ways of solving the same problem. In this particular question, you have to find the values of T1 and T2.

In the part I'm on now, you solve it by drawing a Triangle of Force.

I understand how to do the Vectors, but I don't know how they've worked out the angles in the triangle. Could anyone manage to explain this to me at all?

I have a picture I've drawn on "Paint" of the original 3 forces, plus the triangle of force, and I've tried to paste it here, but it doesn't seem to be working, and I'm not sure how to do it.

To describe it, the force diagram consists of an object of mass 12 Kg,suspended by 2 strings called T1 and T2. String T1 is sort of pointing upwards and leftwards, forming an angle of 27 degrees with the horizon at the top. String T2 is pointing upwards and rightwards, forming an angle of 38 degrees with the horizon at the top. At the bottom of the 2 strings, where they meet, hangs the object of 12gN.

I'm sorry I can't manage to get the diagram put up here. I'll have to try and learn how to do it.
Oh, actually I think I have managed to upload a picture of the original forces. Now I'm going to try and attach a picture of their triangle of forces.

IN THE TRIANGLE THAT I CAN NOW SEE UNDER MY POST, THE ANGLES IN THEIR FORCE TRIANGLE ARE:
BOTTOM LEFT - 52 degrees
RIGHT - 65 degrees
TOP - 63 degree
The vectors in the triangle of force are - VERTICAL ONE ON LEFT = 12g. ONE AT TOP = T1. ONE STARTING AT BOTTOM AND SLOPING UPWARDS AND RIGHTWARDS = T2[/U]
[IMG]file:///C:/Users/Moi/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]
[IMG]file:///C:/Users/Moi/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]
[IMG]file:///C:/Users/Moi/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg[/IMG]
Resolve the forces vertically:

$T_1 sin(27^{\circ})$
$T_2 sin(38^{\circ})$

If the object is not moving then the resultant force must be equal to 0.

$T_1 sin(27^{\circ}) + T_2 sin(38^{\circ}) = 12g$

Resolving the forces horizontally:

$T_1 \cos(27^{\circ})$
$T_2 \cos(38^{\circ})$
As there is no horizontal force on the mass then that is equal to 0

Again, as there is no overall force these two are equal

$T_1 \cos(27^{\circ}) = T_2 \cos(38^{\circ})$

To find T1 and T2 solve the simultaneous equations

3. ## it's the method of deciding the angles that I need

Originally Posted by e^(i*pi)
Resolve the forces vertically:

$T_1 sin(27^{\circ})$
$T_2 sin(38^{\circ})$

If the object is not moving then the resultant force must be equal to 0.

$T_1 sin(27^{\circ}) + T_2 sin(38^{\circ}) = 12g$

Resolving the forces horizontally:

$T_1 \cos(27^{\circ})$
$T_2 \cos(38^{\circ})$
As there is no horizontal force on the mass then that is equal to 0

Again, as there is no overall force these two are equal

$T_1 \cos(27^{\circ}) = T_2 \cos(38^{\circ})$

To find T1 and T2 solve the simultaneous equations
hello E ...pi,

Thanks very much for your reply, which I have read through carefully, but really it was a kind of revision & practice for me of something I've done already, helpful and clarifying ,such as reading the statements about things being equal to zero etc, thanks, but the thing I am stuck on is how they find the angles to put into the triangle of force.

I'm familiar with things like looking for supplementary angles, moving the vectors in your mind's eye etc, angles on a straight line adding up to 180 etc, but I can't seem to manage to see how to apply any of those methods here, and I would be grateful for any suggestion.

4. Originally Posted by protractor
hello E ...pi,

Thanks very much for your reply, which I have read through carefully, but really it was a kind of revision & practice for me of something I've done already, helpful and clarifying ,such as reading the statements about things being equal to zero etc, thanks, but the thing I am stuck on is how they find the angles to put into the triangle of force.

I'm familiar with things like looking for supplementary angles, moving the vectors in your mind's eye etc, angles on a straight line adding up to 180 etc, but I can't seem to manage to see how to apply any of those methods here, and I would be grateful for any suggestion.
My apologies, I thought you'd used it as a method to solve the equation.

The force triangle comes because you can move vectors as long as you don't change the angle or the magnitude.

From what I can gather off your second diagram the vector T2 has been moved back so that the end is at the intersection of the mass and T1. Since it's still pointing in the same direction we do not reverse it's sign

From there the mg force has been moved across to form a perpendicular with the horizon. This means that the angle between mg and T1 must be equal to 90 degrees and hence the angle is 90-27 = 63

If you extend the original line of mg up to the horizon you it will form a right triangle with T2 as the hypotenuse. As all the angles in a triangle must add up to 180 degrees the missing angle is 52. When the vector T2 is moved the angle stays the same.

Again using the fact that all angles in a triangle add up to 180 we get the last angle

edit: see attachment

5. ## movements

Originally Posted by e^(i*pi)
My apologies, I thought you'd used it as a method to solve the equation.

The force triangle comes because you can move vectors as long as you don't change the angle or the magnitude.

From what I can gather off your second diagram the vector T2 has been moved back so that the end is at the intersection of the mass and T1. Since it's still pointing in the same direction we do not reverse it's sign

From there the mg force has been moved across to form a perpendicular with the horizon. This means that the angle between mg and T1 must be equal to 90 degrees and hence the angle is 90-27 = 63

If you extend the original line of mg up to the horizon you it will form a right triangle with T2 as the hypotenuse. As all the angles in a triangle
must add up to 180 degrees the missing angle is 52. When the vector T2 is moved the angle stays the same.

Again using the fact that all angles in a triangle add up to 180 we get the last angle

edit: see attachment
Oh thanks E..pi, that's really helpful, I like the way you can extend the line of mg up to the horizon, how interesting.