Moment about a point using graph paper

[werson tan]

Homework Statement

the ans given is 95Nm , but i gt 80.9Nm , which part i did wrongly ?

Homework Equations

The Attempt at a Solution

20(4) -39(5/surd 26(3)) +39(1/surd26)(1) -60(3/5)(1)+60(4/5)(3)
=80.9[/B]

 

[BvU]

What is a surd26 ?
Could you tidy up your notation in general as well. 20(4) probably means 20 N * 4 m = 80 Nm ?

[edit] sorry, that 20(4) is a correct notation. Just had to get used to the sight.

 

[mjsd]

firstly, I don't think

20(4) -39(5/surd 26(3)) +39(1/surd26)(1) -60(3/5)(1)+60(4/5)(3)

would yield 80.9

The term 39(1/ Sqrt[26])(1) should not be there... 60 (4/5)(3) the (3) should be a (2)

however with these adjustments, the answer is still not 95 Nm clockwise by my calculations (more like 25-ish), I believe there might be typo somewhere??

 

[BvU]

Sigh... after wasting some time on this, I conclude that surd26 is $\sqrt { 26}$ and you use it to decompose the 39 N into x and y components. For the $\ \vec r\$ in $\ \vec \tau = \vec r \times \vec F\$ you then take $\ \vec r = (1,3)\$. Shouldn't that be $\ \vec r = (0,3) \$ ? It gives yet another answer, but I wouldn't trust a book answer in a book that let's 1 m be represented by a square ( )

PS I get what mjsd gets. The bat types faster ...

 

[werson tan]

Sigh... after wasting some time on this, I conclude that surd26 is $\sqrt { 26}$ and you use it to decompose the 39 N into x and y components. For the $\ \vec r\$ in $\ \vec \tau = \vec r \times \vec F\$ you then take $\ \vec r = (1,3)\$. Shouldn't that be $\ \vec r = (0,3) \$ ? It gives yet another answer, but I wouldn't trust a book answer in a book that let's 1 m be represented by a square ( )

PS I get what mjsd gets. The bat types faster ...

sorry , the book contain error , the 1m should be 12m on the 39N triangle . I have redo the question .
20(4) -39(5/13)(3) +39(12/13)(1) -60(3/5)(1)+60(4/5)(3)
=179N

 

[mjsd]

sorry , the book contain error , the 1m should be 12m on the 39N triangle . I have redo the question .
20(4) -39(5/13)(3) +39(12/13)(1) -60(3/5)(1)+60(4/5)(3)
=179N

As I said in my first post...
39(12/13)(1) should NOT be there as the (1) is actually (0)...can you see that?
and
60(4/5)(3) is actually 60(4/5)(2)
once you have got these
you should get 95 Nm for your answers

 

[werson tan]

As I said in my first post...
39(12/13)(1) should NOT be there as the (1) is actually (0)...can you see that?
and
60(4/5)(3) is actually 60(4/5)(2)
once you have got these
you should get 95 Nm for your answers

I know what u mean . In the above steps , I have resolved the 60N in this way( black and red) , whereas the green colour one is the resolution of force done by you .

For the 39N force , I have resolved the force in this way ( black and red)

 

[werson tan]

I know what u mean . In the above steps , I have resolved the 60N and 39N in this way( black and red) , whereas the green colour one is the resolution of force done by you .

For the 39N force , I have resolved the force in this way ( black and red)

yes , i know why it should be 39(12/13)(0), now my question is why can't I resolve the force in the way that i have posted earlier?
I have resolved the 60N and 39N in this way( black and red) , whereas the green colour one is the resolution of force done by you .