M1 Moments

Original post by Steelmeat

T=\dfrac{W}{3}

xW+2(

\frac{W}{3}

)=0
xW+

\frac{2}{3}w

=0
x=

-\dfrac{2}{3}

so does that mean that actually x is

\frac{2}{3}

m to the left of C?

You should get

\frac{2}{3}

but it should be from B and non-negative.

Reply 2

Original post by Cryptokyo

You should get

\frac{2}{3}

but it should be from B and non-negative.

The reason for this is that your equations should be

xW-2\left(\frac{W}{3}\right)=0

as moments act in different directions.

As the sum of the anticlockwise moments = sum of clockwise moments.

(edited 7 years ago)

Reply 3

Original post by Cryptokyo

You should get

\frac{2}{3}

but it should be from B and non-negative.

i don't quite understand? i've subbed the equation for t in terms of w back into the equation where i took moments about C

Reply 4

Your walking is all correct apart from the sign errors in the moments.

Reply 5

Original post by Steelmeat

i don't quite understand? i've subbed the equation for t in terms of w back into the equation where i took moments about C

When you take moments about a point the anticlockwise moments are equal to the clockwise moments. The weight of the rod acts clockwise and the tension in the string acts anticlockwise when moments are taken about B. And the inverse is true for moments about C.

Reply 6

Hence the moments about B should be

xW-2T=0

and about C should be

T-(2-x)W=0

Reply 7

Original post by Cryptokyo

yea

\sum of\ anticlockwise\ moments = \sum of\ clockwise\ moments

taking moments about B weight w acts clockwise = tension in the string which is anticlockwise

.... so i'm right then?

Reply 8

Original post by Steelmeat

yea

\sum of\ anticlockwise\ moments = \sum of\ clockwise\ moments

taking moments about B weight w acts clockwise = tension in the string which is anticlockwise

.... so i'm right then?

You are correct in way but your answer is negative as your equations give

x=-\frac{2}{3}

due to sign errors in the moments it should be that

xW=2T

and that

T=\frac{W}{3}

Reply 9

Original post by Cryptokyo

Hence the moments about B should be

xW-2T=0

and about C should be

T-(2-x)W=0

?????????

no when you take moments about something you don't include the moment about itself....

Reply 10

@Zacken sort me out fam

Reply 11

Original post by Steelmeat

@Zacken sort me out fam

Original post by Steelmeat

Taking moments about C W*x+2*T=0
2T+xW=0

Why are you saying that tension and weight are acting in the same direction?

Reply 12

Original post by Zacken

Why are you saying that tension and weight are acting in the same direction?

Yes they both act clockwise, i put the little arrow acting up the string for tension is that wrong?

Reply 13

Original post by Steelmeat

Yes they both act clockwise, i put the little arrow acting up the string for tension is that wrong?

So tension acts vertically upwards and weight acts vertically downwards but they both produce a clockwise moment... what?

Reply 14

Original post by Zacken

So tension acts vertically upwards and weight acts vertically downwards but they both produce a clockwise moment... what?

yes lemme post a pic of my diagram

Edit: i'm gonna redraw it since it's a bit messy

Reply 15

Original post by Steelmeat

yes lemme post a pic of my diagram

Edit: i'm gonna redraw it since it's a bit messy

Don't bother dude. Put your rule parallel to the plank. Now push push the ruler vertically upwards, does it turn anticlockwise or clockwise?

Now push the ruler vertically downwards, does it turn anticlockwise or clockwise?

Reply 16

Original post by Zacken

pushing on the left hand side of the ruler pushing upwards turns the ruler clockwise

pushing vertically down on the right hand side of the ruler turns it clockwise also

Reply 17

Original post by Steelmeat

pushing on the left hand side of the ruler pushing upwards turns the ruler clockwise

pushing vertically down on the right hand side of the ruler turns it clockwise also

Oh, you've put the weight to the right of C. Urgh.

So that means that the centre of mass is 2/3 to the left of C, yes.

So the distance from A is 1 + 2 - (2/3) = 7/3.

Reply 18