Solomon paper f

Figure 2 shows 4 points A, B, C and D arranged such that they form the corners of a square of
side 2 m. Forces of 5 N, 3 N, 2 N and 4 N act in the directions AB , BC , DC and DA
respectively.
(a) Calculate the magnitude and sense of the resultant moment about A. (3 marks)
An additional force of magnitude X Newtons is added in the direction CA . The resultant
moment of all the forces about D is now zero.
(b) Find, in the form k√2, the value of X.

i got part a but on part b i keep getting the distance of X from d to be root 2 not 2root2 which is what it says on the mark scheme?
any ideas?

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

Are you sure you've posted in the right place? :smile:

Here's a link to our subject forum which should help get you more responses if you post there. :redface:

You can also find the Exam Thread list for A-levels here and GCSE here. :dumbells:

Just quoting in Puddles the Monkey so she can move the thread if needed :h:

Spoiler

Reply 2

Zacken

Original post by Lola1244

Could you show us a picture of your working or a diagram?

Reply 3

Lola1244

Original post by Zacken

Could you show us a picture of your working or a diagram?

i just dont get how they have calculated the perpendicular distance????

Reply 4

Dominator1

They've calculated the perpendicular distance using pythagoras theorem to give square root of eight which equals 2 root 2. Now there are two forces with a total of 16Nm acting anticlockwise of D, so there has to be 16Nm acting clockise. so
2√2*F=16
F=16/2√2
F=8/√2
F=4√2 N

Reply 5

coconut64

Original post by Dominator1

Can you explain how to do part a? Thanks.

Reply 6

coconut64

Original post by Lola1244

i just dont get how they have calculated the perpendicular distance????

Hey could you please explain with you did part a? Did you work out the diagonal distance? I don't get what you really need to work out... Thanks!

Reply 7

Cryptokyo

Original post by coconut64

Hey could you please explain with you did part a? Did you work out the diagonal distance? I don't get what you really need to work out... Thanks!

This was done by taking moments about A. But the key point is that forces that act through the point that you are taking moments about have a moment of 0Nm. Thus in the diagram the 4N force and the 5N force act through A so each have a moment of 0. However, the 2N and the 3N forces have moments. The 3N force acts anticlockwise with moment 3x2=6Nm and the 2N force acts clockwise and has moment 2x2=4Nm. Therefore the resultant moment is 6-4 = 2Nm

Reply 8

coconut64

Original post by Cryptokyo

What distance are you using because I don't get how you can work out the distance ( 2m)? Where does it come from ? Thanks. ps I get why only two forces are involved .

(edited 7 years ago)

Reply 9

Dominator1

Original post by coconut64

What distance are you using because I don't get how you can work out the distance ( 2m)? Where does it come from ? Thanks. ps I get why only two forces are involved .

The distance is given in the question.

Reply 10

coconut64

Original post by Dominator1

The distance is given in the question.

OH so each side would just be 2m. But how do you know whether the force acts in the clockwise direction or anti clockwise direction? Thanks.

Reply 11

Dominator1

Original post by coconut64

OH so each side would just be 2m. But how do you know whether the force acts in the clockwise direction or anti clockwise direction? Thanks.

Imagine that you have this as a wooden square in front of you. Now put a nail through point A (only point A) so you cannot move that edge. Let four of your freinds apply the forces shown in the question by their fingertips in the directions shown. Once you visualise that you'll get it.

Reply 12

intesar

How did you get the perpendicular distance as 2√2? If you do sqrt(2^2+2^2)=2√2 then you have to half it because that's double the perpendicular distance. I think it would be sqrt 2. I will need to ask my teacher but it could be a mark scheme error.