Help with equationss

[Fiona G]

Hi, I am 15yrs old and I have just started taking engineering science. I am struggling to understand how to solve these equations and I was wondering if someone could help explain them in really simple terms?

Please note, I am NOT asking for the answers; I want to learn how to do these myself, so I would really love a breakdown of what to do and why we’re doing it.

Thanks in advance.

[Sensei]

Did you already have Pythagoras' theorem of triangles in math?

 

[studiot]

Hello, Fiona and welcome.

You say you are just starting Engineering Science, but in the UK we are on school/college holidays.

So are you trying to get ahead or to catch up these holidays ?

Either way you have come to the right place.

 

So your question is from the Mechanics (statics) part of Engineering Science and refers to equilibrium.

You need to know the three Laws of static equilibrium

1) The sum of all vertical forces equals zero.

2) The sum of all horizontal forces equals zero.

3) The sum of all clockwise moments equals the sum of all anticlockwise moments.

 

You will not need all of these three for these early questions, but you will need some very simple trigonometry and how to draw and understand the 'free body diagram' on the right of your top diagram.

We should start with this diagram so can you say why it has been drawn as it has and what it means ?

If not then we can work through this first.

 

Edit the site system has merged my two replies.

I see you came back to look so here is a small hint.

The log is in horizontal equilibrium.  (Law 2)

That means that it is neither moving to the left or to the right.

That means that the sum of the lefward forces = the sum of the righward forces     or  the total sum of the horizontal forces = 0

There is only one leftward force ( the horizontal part of 8.17kN) and only one rightward force  (the horizontal part of 10.5 kN)

 

Note is say the horizpontal part because each force is acting at an angle to the horizontal.

The value of the horizontal part is given by the formula

horizontal force = actual force times the cosine of the angle the line of the force makes with the horizontal.

 

Please check for yourself that

8.17 * cos (50)  = 10.6  *  cos (60)

 

Is any of this familiar ?

Edited August 26, 2023 by studiot

[Fiona G]

17 hours ago, Sensei said:

Did you already have Pythagoras' theorem of triangles in math?

 

Hi, yes I do understand Pythagoras  I’m just not sure how I should apply it to the second question?

[Genady]

The OP is asking about

18 hours ago, Fiona G said:

these equations

but there are no equations.

[Fiona G]

16 hours ago, studiot said:

Hello, Fiona and welcome.

You say you are just starting Engineering Science, but in the UK we are on school/college holidays.

So are you trying to get ahead or to catch up these holidays ?

Either way you have come to the right place.

 

So your question is from the Mechanics (statics) part of Engineering Science and refers to equilibrium.

You need to know the three Laws of static equilibrium

1) The sum of all vertical forces equals zero.

2) The sum of all horizontal forces equals zero.

3) The sum of all clockwise moments equals the sum of all anticlockwise moments.

 

You will not need all of these three for these early questions, but you will need some very simple trigonometry and how to draw and understand the 'free body diagram' on the right of your top diagram.

We should start with this diagram so can you say why it has been drawn as it has and what it means ?

If not then we can work through this first.

 

Edit the site system has merged my two replies.

I see you came back to look so here is a small hint.

The log is in horizontal equilibrium.  (Law 2)

That means that it is neither moving to the left or to the right.

That means that the sum of the lefward forces = the sum of the righward forces     or  the total sum of the horizontal forces = 0

There is only one leftward force ( the horizontal part of 8.17kN) and only one rightward force  (the horizontal part of 10.5 kN)

 

Note is say the horizpontal part because each force is acting at an angle to the horizontal.

The value of the horizontal part is given by the formula

horizontal force = actual force times the cosine of the angle the line of the force makes with the horizontal.

 

Please check for yourself that

8.17 * cos (50)  = 10.6  *  cos (60)

 

Is any of this familiar ?

Hi, thanks for your reply. 

I’m in Scotland, and school started for us on 16th August.

I understand the laws of static equilibrium, and I’m also familiar with FBDs.

 

4 minutes ago, Genady said:

The OP is asking about

but there are no equations.

Yes, you’re right. I meant questions. My apologies. 

[studiot]

11 minutes ago, Fiona G said:

16 hours ago, studiot said:

Expand  

Hi, thanks for your reply. 

I’m in Scotland, and school started for us on 16th August.

I understand the laws of static equilibrium, and I’m also familiar with FBDs.

Great stuff.

 

So did you do my little sum for me ?

 

 

[Fiona G]

16 hours ago, studiot said:

Hello, Fiona and welcome.

You say you are just starting Engineering Science, but in the UK we are on school/college holidays.

So are you trying to get ahead or to catch up these holidays ?

Either way you have come to the right place.

 

So your question is from the Mechanics (statics) part of Engineering Science and refers to equilibrium.

You need to know the three Laws of static equilibrium

1) The sum of all vertical forces equals zero.

2) The sum of all horizontal forces equals zero.

3) The sum of all clockwise moments equals the sum of all anticlockwise moments.

 

You will not need all of these three for these early questions, but you will need some very simple trigonometry and how to draw and understand the 'free body diagram' on the right of your top diagram.

We should start with this diagram so can you say why it has been drawn as it has and what it means ?

If not then we can work through this first.

 

Edit the site system has merged my two replies.

I see you came back to look so here is a small hint.

The log is in horizontal equilibrium.  (Law 2)

That means that it is neither moving to the left or to the right.

That means that the sum of the lefward forces = the sum of the righward forces     or  the total sum of the horizontal forces = 0

There is only one leftward force ( the horizontal part of 8.17kN) and only one rightward force  (the horizontal part of 10.5 kN)

 

Note is say the horizpontal part because each force is acting at an angle to the horizontal.

The value of the horizontal part is given by the formula

horizontal force = actual force times the cosine of the angle the line of the force makes with the horizontal.

 

Please check for yourself that

8.17 * cos (50)  = 10.6  *  cos (60)

 

Is any of this familiar ?

So I got as far as:

8.17kN * cos (50) = 10.5kN * cos (60)

8170 * cos (50) = 10500n * cos (60)

5251.57n + 5250n = weight 10501.57n

 

but I’m fairly confident that’s wrong…

[studiot]

29 minutes ago, Fiona G said:

So I got as far as:

8.17kN * cos (50) = 10.5kN * cos (60)

8170 * cos (50) = 10500n * cos (60)

5251.57n + 5250n = weight 10501.57n

 

but I’m fairly confident that’s wrong…

The first part is correct

8.17kN * cos (50) = 10.5kN * cos (60)  = 5.25 kN numerically

But the correct equation from Law 2 is that they are equal and opposite since they are acting in opposite directions !

So to sum to zero one must be positive and one must be negative.

So we have, taking left to right as the positive direction horizontally

10.5kN * cos (60) +  ( - (8.17kN * cos (50) ) = 0

 

However that is the equation for horizontal equilibrium.

 

For the log weight, you need the equation for vertical equilibrium of the log.

 

Using what I have just shown you and taking up as positive,  can you write the vertical equation ?

 

Edited August 27, 2023 by studiot

[Fiona G]

 

1 hour ago, studiot said:

The first part is correct

8.17kN * cos (50) = 10.5kN * cos (60)  = 5.25 kN numerically

But the correct equation from Law 2 is that they are equal and opposite since they are acting in opposite directions !

So to sum to zero one must be positive and one must be negative.

So we have, taking left to right as the positive direction horizontally

10.5kN * cos (60) +  ( - (8.17kN * cos (50) ) = 0

 

However that is the equation for horizontal equilibrium.

 

For the log weight, you need the equation for vertical equilibrium of the log.

 

Using what I have just shown you and taking up as positive,  can you write the vertical equation ?

 

So would it be:

10.5kN * sin(60) +  ( - (8.17kN * sin (50) ) = 0

I am not too sure, however this is my best guess on how to calculate the verticals.

[studiot]

1 hour ago, Fiona G said:

 

So would it be:

10.5kN * sin(60) +  ( - (8.17kN * sin (50) ) = 0

I am not too sure, however this is my best guess on how to calculate the verticals.

It is very important to always consider both magnitude and direction for forces.

Horizontally the two slings are pulling in opposite directions.

That is why there is a minus sign on one of them.

Do you think they are also pulling in opposite directions vertically ?

 

Quote

So would it be:

10.5kN * sin(60) +  ( - (8.17kN * sin (50) ) = 0

But well done for recognising the change from cosine to sine to get the vertical components (do you understand we are 'resolving vertically and horizontally into two separate components for each force ?)

🙂

Now we must include all forces that are acting on the free body  - the log.

You have forgotten one, the one you are trying to find  -  W.

So you will have three terms in your equation, not two.

 

Fear not it is good to work through at this level of detail as I feel you are learning and using what you already know.

Can you have another go at the vertical equilibrium equation ?

I note you have only been a member for 21 hours.

New members are restricted to 5 posts in their first 24 hours, after that posting is unrestricted.

 

We do this because it prevents ill intententioned new members or bots to post pages and pages of spam, as happens to some forums.

 

Edit

Since I see you are here now I will post the vertical equation for you to compare with yours and solve and speed things along.

 

As the log is under vertical equilibrium by the actions of 3 forces

 

Downward forces = upward forces

W =  10.5 sin60  + 8.17 sin 50

Edited August 27, 2023 by studiot