# Mechanics Resultant of Three Forces and Angle Between the Line of Action Help

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Hello, I have been revising mechanics problems and have come across the questions below which I am having some difficulty with; especially part 3.

I have answered all aspects but would greatly appreciate any guidance or amendments to improve upon my workings and solutions. ๐

Three forces, F1,F2 and F3 act on a particle.

F1=(-6i+2j)

F2=(2i-3j)

F3=(pi+qj)

1. The particle is in a state of equilibrium, determine the values of p and q.

As the particle is in equilibrium the algebraic sum of the components in each direction will be zero.

(-6i+2j)+(2i-3j)+(pi+qj)=0

Factoring;

i(2-6+p)+j(2-3+q)=0

Since the resultant is zero, the coefficient of i is zero:

2-6+p=0

p=4

Since the resultant is zero, the coefficient of j is also zero:

2-3+q=0

q=1

The resultant of the forces F1 and F2 is R

2. Calculate in Newtons the magnitude of R

I understand that one can use Pythagoras's Theorem to calculate R.

โฃRโฃ=sqrt((-6i+2j)^2+(2i-3j)^2)

โฃRโฃ=sqrt((-4)^2+(-1)^2)

โฃRโฃ=sqrt(17)

โฃRโฃ=4.12 N to 3.s.f.

3. Calculate, to the nearest degree, the angle between the line of action of R and the vector i.

This is where I am having the greatest trouble and I have attached a diagram to exhibit my workings.

If ฮธ is the angle between R and the vector i;

tan ฮธ= opp/adj

tan ฮธ=1/4

ฮธ=tan^-1(1/4)

ฮธ=14 degrees to the nearest degree

Thank you to anyone who replies ๐

I have answered all aspects but would greatly appreciate any guidance or amendments to improve upon my workings and solutions. ๐

Three forces, F1,F2 and F3 act on a particle.

F1=(-6i+2j)

F2=(2i-3j)

F3=(pi+qj)

1. The particle is in a state of equilibrium, determine the values of p and q.

As the particle is in equilibrium the algebraic sum of the components in each direction will be zero.

(-6i+2j)+(2i-3j)+(pi+qj)=0

Factoring;

i(2-6+p)+j(2-3+q)=0

Since the resultant is zero, the coefficient of i is zero:

2-6+p=0

p=4

Since the resultant is zero, the coefficient of j is also zero:

2-3+q=0

q=1

The resultant of the forces F1 and F2 is R

2. Calculate in Newtons the magnitude of R

I understand that one can use Pythagoras's Theorem to calculate R.

โฃRโฃ=sqrt((-6i+2j)^2+(2i-3j)^2)

โฃRโฃ=sqrt((-4)^2+(-1)^2)

โฃRโฃ=sqrt(17)

โฃRโฃ=4.12 N to 3.s.f.

3. Calculate, to the nearest degree, the angle between the line of action of R and the vector i.

This is where I am having the greatest trouble and I have attached a diagram to exhibit my workings.

If ฮธ is the angle between R and the vector i;

tan ฮธ= opp/adj

tan ฮธ=1/4

ฮธ=tan^-1(1/4)

ฮธ=14 degrees to the nearest degree

Thank you to anyone who replies ๐

Last edited by Alexandramartis; 1 month ago

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#2

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Hello, I have been revising mechanics problems and have come across the questions below which I am having some difficulty with; especially part 3.

I have answered all aspects but would greatly appreciate any guidance or amendments to improve upon my workings and solutions. ๐

Three forces, F1,F2 and F3 act on a particle.

F1=(-6i+2j)

F2=(2i-3j)

F3=(pi+qj)

1. The particle is in a state of equilibrium, determine the values of p and q.

As the particle is in equilibrium the algebraic sum of the components in each direction will be zero.

(-6i+2j)+(2i-3j)+(pi+qj)=0

Factoring;

i(2-6+p)+j(2-3+q)=0

Since the resultant is zero, the coefficient of i is zero:

2-6+p=0

p=4

Since the resultant is zero, the coefficient of j is also zero:

2-3+q=0

q=1

The resultant of the forces F1 and F2 is R

2. Calculate in Newtons the magnitude of R

I understand that one can use Pythagoras's Theorem to calculate R.

โฃRโฃ=sqrt((-6i+2j)^2+(2i-3j)^2)

โฃRโฃ=sqrt((-4)^2+(-1)^2)

โฃRโฃ=sqrt(17)

โฃRโฃ=4.12 N to 3.s.f.

3. Calculate, to the nearest degree, the angle between the line of action of R and the vector i.

This is where I am having the greatest trouble and I have attached a diagram to exhibit my workings.

If ฮธ is the angle between R and the vector i;

tan ฮธ= opp/adj

tan ฮธ=1/4

ฮธ=tan^-1(1/4)

ฮธ=14 degrees to the nearest degree

Thank you to anyone who replies ๐

**Alexandramartis**)Hello, I have been revising mechanics problems and have come across the questions below which I am having some difficulty with; especially part 3.

I have answered all aspects but would greatly appreciate any guidance or amendments to improve upon my workings and solutions. ๐

Three forces, F1,F2 and F3 act on a particle.

F1=(-6i+2j)

F2=(2i-3j)

F3=(pi+qj)

1. The particle is in a state of equilibrium, determine the values of p and q.

As the particle is in equilibrium the algebraic sum of the components in each direction will be zero.

(-6i+2j)+(2i-3j)+(pi+qj)=0

Factoring;

i(2-6+p)+j(2-3+q)=0

Since the resultant is zero, the coefficient of i is zero:

2-6+p=0

p=4

Since the resultant is zero, the coefficient of j is also zero:

2-3+q=0

q=1

The resultant of the forces F1 and F2 is R

2. Calculate in Newtons the magnitude of R

I understand that one can use Pythagoras's Theorem to calculate R.

โฃRโฃ=sqrt((-6i+2j)^2+(2i-3j)^2)

โฃRโฃ=sqrt((-4)^2+(-1)^2)

โฃRโฃ=sqrt(17)

โฃRโฃ=4.12 N to 3.s.f.

3. Calculate, to the nearest degree, the angle between the line of action of R and the vector i.

This is where I am having the greatest trouble and I have attached a diagram to exhibit my workings.

If ฮธ is the angle between R and the vector i;

tan ฮธ= opp/adj

tan ฮธ=1/4

ฮธ=tan^-1(1/4)

ฮธ=14 degrees to the nearest degree

Thank you to anyone who replies ๐

Q2 is the right answer, but the first line does not match the answer.

Q3 is right, but 180 out. Draw Fs & R.

Last edited by mqb2766; 1 month ago

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Thank you for your reply, yes I see what you mean about question 2 I will amend that;

โฃRโฃ=sqrt((-6i+2i)^2+(2j-3j)^2)

โฃRโฃ=sqrt((-4)^2+(-1)^2)

โฃRโฃ=sqrt(17)

โฃRโฃ=4.12 N to 3.s.f.

3. I am sorry I do not undertstand what you mean here?

Thank you again ๐

โฃRโฃ=sqrt((-6i+2i)^2+(2j-3j)^2)

โฃRโฃ=sqrt((-4)^2+(-1)^2)

โฃRโฃ=sqrt(17)

โฃRโฃ=4.12 N to 3.s.f.

3. I am sorry I do not undertstand what you mean here?

Thank you again ๐

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#4

(Original post by

Thank you for your reply, yes I see what you mean about question 2 I will amend that;

โฃRโฃ=sqrt((-6i+2i)^2+(2j-3j)^2)

โฃRโฃ=sqrt((-4)^2+(-1)^2)

โฃRโฃ=sqrt(17)

โฃRโฃ=4.12 N to 3.s.f.

3. I am sorry I do not undertstand what you mean here?

Thank you again ๐

**Alexandramartis**)Thank you for your reply, yes I see what you mean about question 2 I will amend that;

โฃRโฃ=sqrt((-6i+2i)^2+(2j-3j)^2)

โฃRโฃ=sqrt((-4)^2+(-1)^2)

โฃRโฃ=sqrt(17)

โฃRโฃ=4.12 N to 3.s.f.

3. I am sorry I do not undertstand what you mean here?

Thank you again ๐

For Q2, I'd drop the i and j when putting into the pythagoras formula.

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Thank you I will take your advice for question 2.

I will draw another diagram and upload it now... ๐

I will draw another diagram and upload it now... ๐

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(Original post by

You need to draw F1, F2 and R, or do the algebra correctly, or both. Upload the corrected sketch?

For Q2, I'd drop the i and j when putting into the pythagoras formula.

**mqb2766**)You need to draw F1, F2 and R, or do the algebra correctly, or both. Upload the corrected sketch?

For Q2, I'd drop the i and j when putting into the pythagoras formula.

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#7

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Hello, I have drawn another diagram but I have confused myself to a greater extent and do not think that it is correct.

**Alexandramartis**)Hello, I have drawn another diagram but I have confused myself to a greater extent and do not think that it is correct.

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(Original post by

Add F2 to the end of F1 (or vice versa). You know what R is? You didn't state R clearly in your original post (or others), but this is where the mistake lies.

**mqb2766**)Add F2 to the end of F1 (or vice versa). You know what R is? You didn't state R clearly in your original post (or others), but this is where the mistake lies.

Would R=4.12 N, F1=6 and F2=3

As this is not a right-angled triangle could I use the cosine rule to find the angle between F1 and R?

theta=cos^-1b^2+c^2-a^2/2bc

theta=27.2 degrees to 3.s.f

Or have I completely gone on an erroneous tangent ?

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#9

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I have drawn and attached another diagram.

Would R=4.12 N, F1=6 and F2=3

As this is not a right-angled triangle could I use the cosine rule to find the angle between F1 and R?

theta=cos^-1b^2+c^2-a^2/2bc

theta=27.2 degrees to 3.s.f

Or have I completely gone on an erroneous tangent ?

**Alexandramartis**)I have drawn and attached another diagram.

Would R=4.12 N, F1=6 and F2=3

As this is not a right-angled triangle could I use the cosine rule to find the angle between F1 and R?

theta=cos^-1b^2+c^2-a^2/2bc

theta=27.2 degrees to 3.s.f

Or have I completely gone on an erroneous tangent ?

Q2 and 3 are about finding R in polar form (magnitude and phase). You use pythagoras on R to find the magnitude (Q2) and simple trig to find the angle (atan2). You know R in terms of its i and j components, there is no need to use the cos rule.

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(Original post by

In the diagram, there are no axes, no origin and R is the wrong direction. R = ...?

Q2 and 3 are about finding R in polar form (magnitude and phase). You use pythagoras on R to find the magnitude (Q2) and simple trig to find the angle (atan2). You know R in terms of its i and j components, there is no need to use the cos rule.

**mqb2766**)In the diagram, there are no axes, no origin and R is the wrong direction. R = ...?

Q2 and 3 are about finding R in polar form (magnitude and phase). You use pythagoras on R to find the magnitude (Q2) and simple trig to find the angle (atan2). You know R in terms of its i and j components, there is no need to use the cos rule.

And would R=4.12 N as found by previously calculating its magnitude?

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#11

(Original post by

So could I use Pythagoras and trigonmetry to find the anlge between the line of action of R and the vector i? I think I am confused because I do not fully understand what the vector i is? Is this the sum of the i components of F1 and F2? Sorry to keep misunderstanding.

And would R=4.12 N as found by previously calculating its magnitude?

**Alexandramartis**)So could I use Pythagoras and trigonmetry to find the anlge between the line of action of R and the vector i? I think I am confused because I do not fully understand what the vector i is? Is this the sum of the i components of F1 and F2? Sorry to keep misunderstanding.

And would R=4.12 N as found by previously calculating its magnitude?

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(Original post by

What is the vector R in terms of i and j? You've drawn it, you've calculated the magnitude, what don't you understand, what textbook / ... are you using?

**mqb2766**)What is the vector R in terms of i and j? You've drawn it, you've calculated the magnitude, what don't you understand, what textbook / ... are you using?

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#13

(Original post by

The vector R in terms of i and j is -4i-j N. The magnitude of R is 4.12 N to 3.s.f

**Alexandramartis**)The vector R in terms of i and j is -4i-j N. The magnitude of R is 4.12 N to 3.s.f

In the original post, you drew / calculated R = 4i + j.

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(Original post by

Correct, so update the previous diagram with the i and j axes, F1, F2 and R.

In the original post, you drew / calculated R = 4i + j.

**mqb2766**)Correct, so update the previous diagram with the i and j axes, F1, F2 and R.

In the original post, you drew / calculated R = 4i + j.

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#15

(Original post by

So should i and j are in the opposite directions to my original diagram? And could i represent the x-component while j is the y-componet? Then would R also be in the opposite direction?

**Alexandramartis**)So should i and j are in the opposite directions to my original diagram? And could i represent the x-component while j is the y-componet? Then would R also be in the opposite direction?

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(Original post by

What text book are you using?

**mqb2766**)What text book are you using?

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#17

(Original post by

Heinemann Modular Mathematics Mechanics Textbook, but i have not come across a question like part 3 before which is why I am so uncertain in solving it.

**Alexandramartis**)Heinemann Modular Mathematics Mechanics Textbook, but i have not come across a question like part 3 before which is why I am so uncertain in solving it.

It would still help to upload the correct sketch.

Last edited by mqb2766; 1 month ago

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(Original post by

You've never had to find the angle a vector makes with the i axis (x-axis)?

It would still help to upload the correct sketch.

**mqb2766**)You've never had to find the angle a vector makes with the i axis (x-axis)?

It would still help to upload the correct sketch.

Last edited by Alexandramartis; 1 month ago

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#19

(Original post by

At least not in the way the question is phrased, I have not come across an example or exercise in my textbook.

**Alexandramartis**)At least not in the way the question is phrased, I have not come across an example or exercise in my textbook.

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o

Would the attached diagram now be correct?

(Original post by

Its just finding the angle a vector makes with an axis. Correct the previous sketch to get the signs right and draw the unknown angle on there as well as the axes.

**mqb2766**)Its just finding the angle a vector makes with an axis. Correct the previous sketch to get the signs right and draw the unknown angle on there as well as the axes.

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