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 ?