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

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#1
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.
โฃRโฃ=sqrt((-6i+2j)^2+(2i-3j)^2)
โฃRโฃ=sqrt((-4)^2+(-1)^2)
โฃRโฃ=sqrt(17)
โฃRโฃ=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=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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1 month ago
#2
(Original post by 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.
โฃRโฃ=sqrt((-6i+2j)^2+(2i-3j)^2)
โฃRโฃ=sqrt((-4)^2+(-1)^2)
โฃRโฃ=sqrt(17)
โฃRโฃ=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=1/4
ฮธ=tan^-1(1/4)
ฮธ=14 degrees to the nearest degree

Thank you to anyone who replies ๐
Drawing a diagram always helps.
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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#3
Thank you for your reply, yes I see what you mean about question 2 I will amend that;

โฃRโฃ=sqrt((-6i+2i)^2+(2j-3j)^2)
โฃRโฃ=sqrt((-4)^2+(-1)^2)
โฃRโฃ=sqrt(17)
โฃRโฃ=4.12 N to 3.s.f.

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

Thank you again ๐
0
1 month ago
#4
(Original post by Alexandramartis)
Thank you for your reply, yes I see what you mean about question 2 I will amend that;

โฃRโฃ=sqrt((-6i+2i)^2+(2j-3j)^2)
โฃRโฃ=sqrt((-4)^2+(-1)^2)
โฃRโฃ=sqrt(17)
โฃRโฃ=4.12 N to 3.s.f.

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

Thank you again ๐
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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#5
I will draw another diagram and upload it now... ๐
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#6
(Original post by 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.
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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1 month ago
#7
(Original post by Alexandramartis)
Hello, I have drawn another diagram but I have confused myself to a greater extent and do not think that it is correct.
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.
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#8
(Original post by 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.
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 ?
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1 month ago
#9
(Original post by 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 ?
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.
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#10
(Original post by 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.
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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1 month ago
#11
(Original post by 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?
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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#12
(Original post by 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?
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
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1 month ago
#13
(Original post by 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
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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#14
(Original post by 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.
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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1 month ago
#15
(Original post by 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?
What text book are you using?
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#16
(Original post by mqb2766)
What text book are you using?
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.
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1 month ago
#17
(Original post by 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.
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 mqb2766; 1 month ago
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#18
(Original post by 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.
At least not in the way the question is phrased, I have not come across an example or exercise in my textbook. I will upload the diagram in a moment ๐
Last edited by Alexandramartis; 1 month ago
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1 month ago
#19
(Original post by Alexandramartis)
At least not in the way the question is phrased, I have not come across an example or exercise in my textbook.
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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#20
o

(Original post by 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.
Would the attached diagram now be correct?
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