Physics help!

[@d!_†oX!©]

Ohh!! Thanks a lot Nid! The diagram made it so much easier

But hey wait, so what is the answer?

[Vin]

Ohh!! Thanks a lot Nid! The diagram made it so much easier

But hey wait, so what is the answer?

yeah it did.

LOL!!

[nid404]

Ohh!! Thanks a lot Nid! The diagram made it so much easier

But hey wait, so what is the answer?

LOL! A and B are out cause drag's not in the picture.

C is not possible  If upthrust is greater, submarine is flung out of the water 

So it's D.

[@d!_†oX!©]

Thanks once again Captain Nid !

[nid404]

Thanks once again Captain Nid !

At your service 

[elemis]

Cool, I was right

[MasterMath]

I actually think there is a drag force .. but yeah D is the most reasonable one

[@d!_†oX!©]

can someone help me in this question?

[Vin]

can someone help me in this question?

I'm guessing it's C.

[Saladin]

can someone help me in this question?

The answer is C.

The moments about a point is the force times he perpendicular distance.

Use trigonometry to find the length d.

Then you simply times this by 2N, to get the moments.

Then what you have to do is add the moments on the other side is also anti-clockwise.

[@d!_†oX!©]

I am not sure either...there are so many things in my mind right now, each leading to different answer..

1) It can be A:
                    Moment of a force = force X perpendicular distance
                    Moment of a force = 2sin50 X 0.15 = 0.23 Nm
2) It can be C:
                    Two forces acting simultaneously, thus two moments = 2 X 0.23 = 0.46 Nm
3) It can be D:
                    My book (International A/AS level Physics - CIE endorsed) says -

The torque of a couple is the product of one of the forces and the perpendicular distance between the forces

                   Therefore, according to this, Torque = 2 X 0.15 = 0.60 Nm

Fortunately, I don't have any idea about B, or else I'd be dead =.=
Someone please help me!!

[nid404]

can someone help me in this question?

Force X perpendicular distance between the two forces

2sin50  X 0.3= 0.46Nm

C

[@d!_†oX!©]

Oh I just realized, my 1) point is matching the 3) cause as the book says it is the perpendicular answer, so in that way 0.23 should be right.

Engraved, any say about my answers?
Ahana, the book says something else :/ it says only one of the forces..

[Vin]

Oh I just realized, my 1) point is matching the 3) cause as the book says it is the perpendicular answer, so in that way 0.23 should be right.

Engraved, any say about my answers?
Ahana, the book says something else :/ it says only one of the forces..

I saw the book too. Pg. 107. But a different kind of an example is given. What does the markscheme say?

[nid404]

Oh I just realized, my 1) point is matching the 3) cause as the book says it is the perpendicular answer, so in that way 0.23 should be right.

Engraved, any say about my answers?
Ahana, the book says something else :/ it says only one of the forces..

In a couple the magnitude of the force is the same.