Forces on a body with vectors (IWTSE.org)

[TomK]

Homework Statement

The correct answer is [(20/3)g]N.

Relevant Equations

Vectors, forces in equilibrium.

I think I've made a mistake with the vectors. I was trying to get to the same point as the original working (made by IWTSE, not me), but one of the sides of a triangle can't be 0. I would appreciate some help.

 

[Steve4Physics]

I can’t follow your working but see if this helps.

A, B and C are the tops (points of attachment) of each string. They have the same z-coordinate (4) and therefore all lie in the (horizontal) plane z=4.

Can you show AB = BC = AC = 6? Remember the distance between points [itex] (x_1, y_1, z_1)[/itex] and [itex](x_2, y_2, z_2)[/itex] is [itex] √[(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2)]. [/itex]

This means A, B and C form an equilateral triangle in a horizontal plane.

The 3 strings are equal lengths so, by symmetry, the mass (at P) must be vertically below the centre of triangle ABC.

Can you work out the angle, θ, between each string and the vertical using simple geometry/trig’?

By symmetry the 3 tensions are equal. Each has a vertical component Tcosθ. So in the vertical direction [itex] 3Tcosθ = 10g [/itex] which gives T.

EDIT
Please note that when you write (for example) [itex] \vec{AB} = -\vec{OA} + \vec{OB} [/itex] you *cannot* simply put the magnitudes of each vector into the equation. You have to do proper vector addition which takes each vector's direction into account. The simplest method is to separately combine x, y and z components.

 

[TomK]

I can’t follow your working but see if this helps.

A, B and C are the tops (points of attachment) of each string. They have the same z-coordinate (4) and therefore all lie in the (horizontal) plane z=4.

Can you show AB = BC = AC = 6? Remember the distance between points [itex] (x_1, y_1, z_1)[/itex] and [itex](x_2, y_2, z_2)[/itex] is [itex] √[(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2)]. [/itex]

This means A, B and C form an equilateral triangle in a horizontal plane.

The 3 strings are equal lengths so, by symmetry, the mass (at P) must be vertically below the centre of triangle ABC.

Can you work out the angle, θ, between each string and the vertical using simple geometry/trig’?

By symmetry the 3 tensions are equal. Each has a vertical component Tcosθ. So in the vertical direction [itex] 3Tcosθ = 10g [/itex] which gives T.

EDIT
Please note that when you write (for example) [itex] \vec{AB} = -\vec{OA} + \vec{OB} [/itex] you *cannot* simply put the magnitudes of each vector into the equation. You have to do proper vector addition which takes each vector's direction into account. The simplest method is to separately combine x, y and z components.

I see the mistake I made with the vector addition. Now, I can see that the triangle is equilateral. I understand what you said about the mass lying under the centre of said-triangle, and that the vertical tensions of all three strings must equal 10g.

However, I do not know how to use trigonometry/pythagoras from here. I'm struggling to visualise the problem and know what length the sides of the right-angled triangles would be.

 

[Delta2]

1. The triangle ABC is equilateral
2. All 3 strings have the same length
3. All 3 vertices of the triangle A,B,C lie at the plane z=4 which is perpendicular to the z-axis

The above three facts combined can be used to prove that the particle will be exactly below the center O of the triangle ABC, or in other words that the vector ##\vec{OP}## will be parallel to the z-axis (which z axis is perpendicular to the plane of the triangle z=4). So the triangle OPB (and also the triangles OPA and OPC) will be right triangle with the right angle at O. We know the side PB=4(=PA=PC) and also it is standard process to calculate the distance OB (for an equilateral triangle) that is the distance of the center from a vertex. So we can use pythagorean theorem to calculate ##OP=\sqrt{PB^2-OB^2}##. After that we can calculate ##\cos\alpha=\frac{OP}{PB}##. This angle ##\alpha## is the same for each of the strings (the angle that the string make with the side OP). Each vertical component of tension will be ##T\cos\alpha## (from symmetry considerations the tension T of each string is the same) and the sum of them will equal 10g.

 

[Steve4Physics]

I see the mistake I made with the vector addition. Now, I can see that the triangle is equilateral. I understand what you said about the mass lying under the centre of said-triangle, and that the vertical tensions of all three strings must equal 10g.

However, I do not know how to use trigonometry/pythagoras from here. I'm struggling to visualise the problem and know what length the sides of the right-angled triangles would be.

Explanation is a bit long. Carefully work through it if you want.

Step 1.

The tops of the strings, A, B and C, form an equilateral triangle in the horizontal plane.

Use this diagram: https://www.treenshop.com/Treenshop/ArticlesPages/FiguresOfInterest_Article/The%20Equilateral%20Triangle_files/image036.gif

a = AB=BC=AC= 6 units.

Step 2.

If you pick any of the small (right-angled) triangles in the diagram you should see:
(a/2)/R = cos(30)
R = (a/2)/cos(30º)

Step 3.

We know a/2 = 6/2 = 3 and cos(30º) = √3/2 which gives:
R = 3/(√3/2) = 6/√3

Step 4.

Call ‘X’ the triangle’s centre. Point P (where the strings join) is vertically below X.

Remember:
AB=BC=AC= 6 (sides of horizontal triangle)
AP=BP=CP = 4 (lengths of strings)

Do a drawing and label points A, B, C, P and X. It’s an inverted triangular pyramid.

Step 5.

Look carefully at your drawing. Note APX is right-angled triangle. We can draw it separately:

X A

P

Step 6.

Note AX = R (which we’ve just worked out) and AP=4 (length of string). Now we can use Pythagoras to find XP::
AP² = XP² + AX²
4² = XP² + (6/√3)²
XP² = 16 – 36/3 = 16 – 12 = 4
XP = 2

Step 7.

If we let θ = APX then cosθ = XP/AP = 2/4 = 1/2

Note θ is the angle between each string and the vertical so we’re nearly there.

Step 8

Now we can use the formula explained in previous post:
3Tcosθ = 10g

T = 10g/(3cosθ) = 10g/(3*(1/2)) = 20g/3

 

[Chestermiller]

One way of doing this problem is by brute force, assuming that all the tensions are non-zero. Let (x,y,z) be the coordinates of point where the object (assumed a particle) is connected to the 3 strings and let T1, T2, and T3 be the tensions in the 3 strings. So there are 6 unknowns, and 6 equations are required. Three of the equations are for matching the lengths of the 3 strings, using the 3D version of the Pythagorean theorem. The other three equations are for the force balances in the x, y and z directions.

 

[TomK]

Explanation is a bit long. Carefully work through it if you want.

Step 1.

The tops of the strings, A, B and C, form an equilateral triangle in the horizontal plane.

Use this diagram: https://www.treenshop.com/Treenshop/ArticlesPages/FiguresOfInterest_Article/The%20Equilateral%20Triangle_files/image036.gif

a = AB=BC=AC= 6 units.

Step 2.

If you pick any of the small (right-angled) triangles in the diagram you should see:
(a/2)/R = cos(30)
R = (a/2)/cos(30º)

Step 3.

We know a/2 = 6/2 = 3 and cos(30º) = √3/2 which gives:
R = 3/(√3/2) = 6/√3

Step 4.

Call ‘X’ the triangle’s centre. Point P (where the strings join) is vertically below X.

Remember:
AB=BC=AC= 6 (sides of horizontal triangle)
AP=BP=CP = 4 (lengths of strings)

Do a drawing and label points A, B, C, P and X. It’s an inverted triangular pyramid.

Step 5.

Look carefully at your drawing. Note APX is right-angled triangle. We can draw it separately:

X A

P

Step 6.

Note AX = R (which we’ve just worked out) and AP=4 (length of string). Now we can use Pythagoras to find XP::
AP² = XP² + AX²
4² = XP² + (6/√3)²
XP² = 16 – 36/3 = 16 – 12 = 4
XP = 2

Step 7.

If we let θ = APX then cosθ = XP/AP = 2/4 = 1/2

Note θ is the angle between each string and the vertical so we’re nearly there.

Step 8

Now we can use the formula explained in previous post:
3Tcosθ = 10g

T = 10g/(3cosθ) = 10g/(3*(1/2)) = 20g/3

I understand it now. Thank you.