What is the Correct Magnitude of P in a System of Forces?

[werson tan]

Homework Statement

In this question , I'm asked to find the magnitude of P .

Homework Equations

The Attempt at a Solution

Resultant moment about O =
-10cos30 (4√3) -10sin 30 (10
+15sin30(2) + 15cos30(4√3)
-10cos30(4√3) -10sin30(4)
-15cos30(4√3) -15sin30(7)
+10cos30(4√3) +10sin30(11)
+P cos 60 (4√3) =0
P=22N , but the ans given is 15.6N , which part i did wrongly ?

[/B]

 

[BvU]

What happened to $P\,\sin(60^\circ)$ ?
Never mind, P_y is zero.

But what is $4\sqrt 3$ ?

I also wonder what the 1, 2, 4, 7, 11 stand for.

This time I would bet on the book answer (15.58846), even though they draw 30, 45 and 60 degree angles all alike...

 

[werson tan]

Psin(60∘)P\,\sin(60^\circ)

it's parallel with 4surd 3 , so , i didnt take in into calculation .
tan30=4/A
A=4 surd 3

 

[insightful]

I suggest redrawing the sketch roughly to scale. Except for P, all the forces are on parallel lines. Extend those lines and draw a common perpendicular from them through O. All triangles are 30-60-90 and all distances can be gotten without trig using the 1:2:sqrt(3) ratios.

 

[werson tan]

What happened to $P\,\sin(60^\circ)$ ?
Never mind, P_y is zero.

But what is $4\sqrt 3$ ?

I also wonder what the 1, 2, 4, 7, 11 stand for.

This time I would bet on the book answer (15.58846), even though they draw 30, 45 and 60 degree angles all alike...

you gt the ans = 15.5846 ? how do u gt it ? can you help ?

 

[BvU]

Sure, that's what we are for.

it's parallel with 4surd 3 , so , i didnt take in into calculation . (1)
tan30=4/A $\quad$ A=4 surd 3 (2)

(1) I understand ($P_y=0$). But (2) ? Where do you need $4/\tan(30^\circ)$ ?

My question on the 1,2,4,7, 11 was to set you thinking about what these stand for. They are not the x coordinates of the points where the forces act, if that's what you think...

 

[BvU]

For example, the leftmost 10 N clearly grabs to the left of O, so it should have a negative x-coordinate. Idem leftmost 15 N.

Perhaps this also helps you in the other thread (the one on graph paper)

 

[werson tan]

Sure, that's what we are for.
(1) I understand ($P_y=0$). But (2) ? Where do you need $4/\tan(30^\circ)$ ?

My question on the 1,2,4,7, 11 was to set you thinking about what these stand for. They are not the x coordinates of the points where the forces act, if that's what you think...

i gt A = 4/tan30 due to it's a triangle . the horizontal length is 4m, the angle between the P and y-axis is 30 degree , then what does 1, 2, 4, 7, 11 stand for ?

 

[werson tan]

If i am to take the 1, 2, 4, 7, 11 ans the r , then is my working correct ? ( ignore the ans given ) , is my concept correct ?

 

[BvU]

In reply to your post #9:

r has two coordinates, so: No.

In a cartesian coordinate system with x horizontal and y vertical and O at the origin, you want the x coordinate and the y coordinate of the point where the leftmost 10 N acts. etcetera.

----

In reply to your post #8: I asked you first ! (in post #2)

From your post #1:

Resultant moment about O =
-10cos30 (4√3) -10sin 30 (10 $\qquad$ I suppose you mean (1)
+15sin30(2) + 15cos30(4√3)
-10cos30(4√3) -10sin30(4)
-15cos30(4√3) -15sin30(7)
+10cos30(4√3) +10sin30(11)
+P cos 60 (4√3) =0​

The red numbers 1, 2, 4, 7, 11 are wrong. So are all the 4√3

i gt A = 4/tan30 due to it's a triangle . the horizontal length is 4m, the angle between the P and y-axis is 30 degree , then what does 1, 2, 4, 7, 11 stand for ?

I can't find any A in your picture. What do you mean with A ?

(I don't mean 4/tan30, I mean where is it shown in your picture -- which please redraw to scale as insightful also suggested! -- it avoids more than half of your errors !)
​

The angle between P and the y-axis is clearly 90$^\circ$

----

 

[werson tan]

In reply to your post #9:

r has two coordinates, so: No.

In a cartesian coordinate system with x horizontal and y vertical and O at the origin, you want the x coordinate and the y coordinate of the point where the leftmost 10 N acts. etcetera.

----

In reply to your post #8: I asked you first ! (in post #2)

From your post #1:

Resultant moment about O =
-10cos30 (4√3) -10sin 30 (10 $\qquad$ I suppose you mean (1)
+15sin30(2) + 15cos30(4√3)
-10cos30(4√3) -10sin30(4)
-15cos30(4√3) -15sin30(7)
+10cos30(4√3) +10sin30(11)
+P cos 60 (4√3) =0​

The red numbers 1, 2, 4, 7, 11 are wrong. So are all the 4√3

I can't find any A in your picture. What do you mean with A ?

(I don't mean 4/tan30, I mean where is it shown in your picture -- which please redraw to scale as insightful also suggested! -- it avoids more than half of your errors !)
​

The angle between P and the y-axis is clearly 90$^\circ$

----

A is the distance of O to horizontal line where it join all the 15N and 10N

 

[BvU]

Well, you have the wrong value.

 

[werson tan]

Well, you have the wrong value.

Then , wat is the correct one ?

 

[BvU]

For me to know and for you to find out (sorry, that's the PF culture: the answer doesn't help you, you only learn by discovering for yourself). A little hint: it is considerably more than $4\sqrt 3$. Did you make a drawing ?

 

[BvU]

recognize this picture ? If $|\vec A| = 12$, what is $A_x$ and what is $A_y$ ?

 

[werson tan]

recognize this picture ? If $|\vec A| = 12$, what is $A_x$ and what is $A_y$ ?
View attachment 90184

Ax = 12cos60 , Ay = 12 sin 60

 

[BvU]

So A_x = 6 and A_y is ... (not $4\sqrt 3$).

Proceed to list the x-coordinates for the various points where the 10N, 15, 10, 15, 10 N act.

 

[werson tan]

ok , then I'm sure the 1,2,4,7,11 is not correct ? what does those numbers mean ? then , how to determine the perpendicular distance for other forces which is situated at the right of the first 15N and 10N ?

 

[BvU]

Well, you have the y coordinates (at least if you filled in the dots in post #17 -- what did you get ?) and you have the horizontal distances to the leftmost point of the horizontal line where all these 10, 15 N forces act. What is the x coordinate of this leftmost point ?

I figured your post #1 doesn't deal with perpendicular distances but instead evaluates the $\ \vec\tau = \vec r \times \vec F \$ expression ($\ \tau_z = r_x F_y - r_y F_x \$ )​

 

[werson tan]

Well, you have the y coordinates (at least if you filled in the dots in post #17 -- what did you get ?) and you have the horizontal distances to the leftmost point of the horizontal line where all these 10, 15 N forces act. What is the x coordinate of this leftmost point ?

I figured your post #1 doesn't deal with perpendicular distances but instead evaluates the $\ \tau = \vec r \times \vec F \$ expression ($\ \tau_z = r_x F_y - r_y F_x \$ )​

i take the x coordinate of this leftmost point as {0, 0)
isnt the r in the expression ($\ \tau_z = r_x F_y - r_y F_x \$ ) represent perpendicular distance ? why you said my r is not perpendicular distance ?
the y coordinates are all 12sin60= 6 surd(3)

 

[BvU]

i take the x coordinate of this leftmost point as {0, 0)

What, then is the x-coordinate of O ?
$\vec r$ is a position vector $r_x - {\bf O}_x$ is a perpendicular distance only in the x-direction. You can split up the contributions of the force x and y components to the torque this way (as you do in post #1) because the expression is linear.

Good thing you found $6\sqrt 3$ instead of the $4\sqrt 3$. So where are you now with the updated version of post # 1 ?

 

[werson tan]

What, then is the x-coordinate of O ?
$\vec r$ is a position vector $r_x - {\bf O}_x$ is a perpendicular distance only in the x-direction. You can split up the contributions of the force x and y components to the torque this way (as you do in post #1) because the expression is linear.

Good thing you found $6\sqrt 3$ instead of the $4\sqrt 3$. So where are you now with the updated version of post # 1 ?

-10cos30 (6√3) -10sin 30 (1)
+15sin30(2) + 15cos30(6√3)
-10cos30(6√3) -10sin30(4)
-15cos30(6√3) -15sin30(7)
+10cos30(6√3) +10sin30(11)
+P cos 60 (6√3) =0

Now , My P = 18.76N

 

[BvU]

-10cos30 (6√3) -10sin 30 (1)
+ 15cos30(6√3)+15sin30(2)
-10cos30(6√3) -10sin30(4)
-15cos30(6√3) -15sin30(7)
+10cos30(6√3) +10sin30(11)
+P cos 60 (6√3) =0

Now , My P = 18.76N

I hope this isn't driving you crazy ... I admire your resilience and patience.
The red numbers to me seem incorrect on first sight ( a bit pressed for time right now).

Simple check on the first line:

$10\cos{30^\circ}$ N is a force to the left with a perpendicular distance 6√3 so it is a torque that wants to rotate to the left.
$10\sin {30}^\circ$ N is an upwards force that wants to rotate to the right.​

in your summation they want to rotate in the same direction. Can't be right !

But we are getting very close to the correct solution by now, so keep up the good work !

Tip: mark the x and y components of the first 10 N force in the drawing of post #15. $F_x r_y$ is correct (no red there...).
--

 

[werson tan]

removed

 

[BvU]

-10cos30 (6√3) -10sin 30 (3)
+ 15cos30(6√3)+15sin30(2)
-10cos30(6√3) -10sin30(0)
-15cos30(6√3) -15sin30(3)
+10cos30(6√3) +10sin30(7)
+P cos 60 (6√3) =0

"Everything in red" is the answer to your question. The red 0 force does not grab at the same x-coordinate as O_x, even if the bad bad bad picture let's you think so. Remember how you found the $6\sqrt 3$ ? Now find the x-coordinate of the leftmost point of the

horizontal line where it join all the 15N and 10N

From your post #20:

i take the x coordinate of this leftmost point as {0, 0)

You simply can't do that. You already have point O as (0,0)

--

 

[werson tan]

"Everything in red" is the answer to your question. The red 0 force does not grab at the same x-coordinate as O_x, even if the bad bad bad picture let's you think so. Remember how you found the $6\sqrt 3$ ? Now find the x-coordinate of the leftmost point of the

From your post #20:
You simply can't do that. You already have point O as (0,0)

--

-10cos30 (6√3) +10sin 30 (3)
+ 15cos30(2)+15sin30(6√3)
-10cos30(6√3) -10sin30(0)
-15cos30(6√3) -15sin30(3)
+10cos30(6√3) +10sin30(7)
+P cos 60 (6√3) =0
P=14.9N , which part is wrong ?

 

[BvU]

Please don't remove posts. It makes the thread incomprehensible for other readers !

-10cos30 (6√3) -10sin 30 (-3)
+15sin30(6√3) - 15cos30(-2)
-10cos30(6√3) -10sin30(0)
-15cos30(6√3) -15sin30(3)
+10cos30(6√3) +10sin30(7)
+P cos 60 (6√3) =0
P=14.9N , which part is wrong ?

Did you find the x-coordinate of the leftmost point of the "horizontal line where it join all the 15N and 10N" ? (It is not -4 )

 

[werson tan]

Please don't remove posts. It makes the thread incomprehensible for other readers !Did you find the x-coordinate of the leftmost point of the "horizontal line where it join all the 15N and 10N" ? (It is not -4 )

is it (-6,0) ?

 

[BvU]

Yes. The cosine of $\pi/3$ is 0.5

 

[werson tan]

Yes. The cosine of $\pi/3$ is 0.5

how is it related to other forces ?

 

[BvU]

My Yes was premature, sorry. The answer to my ""Did you find the x-coordinate of the leftmost point of the "horizontal line where it join all the 15N and 10N" ? (It is not -4 )"" is not (-6, 0) but -6.

It is a coordinate. It is related to the x coordinates of the points where the 10 and 15 N forces act.

The position vector of this leftmost point is of course $(-6, 6\sqrt 3)$. $\ \ \ (-6,0)$ would be on the x-axis to the left of O.

Can you now calculate the x-coordinate of the point where the leftmost 10 N force is acting ? (it is not 1, it is not -3 and the y-coordinate $6\sqrt 3$ you used is correct.)

And we still have to fix the y-coordinate of the point where the force P is acting. (it is not $6\sqrt 3$ ).

Almost there ..

--

 

[werson tan]

My Yes was premature, sorry. The answer to my ""Did you find the x-coordinate of the leftmost point of the "horizontal line where it join all the 15N and 10N" ? (It is not -4 )"" is not (-6, 0) but -6.

It is a coordinate. It is related to the x coordinates of the points where the 10 and 15 N forces act.

The position vector of this leftmost point is of course $(-6, 6\sqrt 3)$. $\ \ \ (-6,0)$ would be on the x-axis to the left of O.

Can you now calculate the x-coordinate of the point where the leftmost 10 N force is acting ? (it is not 1, it is not -3 and the y-coordinate $6\sqrt 3$ you used is correct.)

And we still have to fix the y-coordinate of the point where the force P is acting. (it is not $6\sqrt 3$ ).

Almost there ..

--

How to calculate the leftmost x coodrinate for 10N? it's shown in the diagram is 3m , right??

 

[BvU]

I really can't understand why you can't make this drawing in more than a week's time. I have no idea what diagram you are talking about. The picture in post #1 is so extremely bad, it deliberately does not show anything as you have been told several times now. You should not use it, except to pick up the given values.

With the diagram you should have made below, it should not be too difficult now, I sincerely hope...

Do yourself a favour, finish the picture and write in the given numbers.

I, in turn, have deliberately misplaced P so that you at least have something left to do yourself to fix the (hopefully last) error in your calculation.

--