# Math Help - some more mechanics help please, (moments)

1. ## some more mechanics help please, (moments)

Can anyone have a look at question 5 on this paper I am really stuck.

i can do 5a but i am stuck on 5b

thanks a lot,

btw i have tried but don't seem to get the right answer,

2. There are 3 forces acting on the beam.

Weight, tension at A, tension at C. You should have gotten

C = 140
A = 70

The beam is in equilibrium, which means no movement. No moments about any point.

Let distance CB = x

Moment about B, from force at C:

$M_c = 140(x)$

Moment about B from force at A:

$M_a = 70(x+90)$

Moment from weight(which acts at the centre):

$M_w = 210(\frac{x+90}{2})$

The sum of these should be zero, and then you have an equation in x to solve!

Remember that anticlockwise moments are positive, and clockwise moments are negative!

3. Originally Posted by Mush
There are 3 forces acting on the beam.

Weight, tension at A, tension at C. You should have gotten

C = 140
A = 70

The beam is in equilibrium, which means no movement. No moments about any point.

Let distance CB = x

Moment about B, from force at C:

$M_c = 140(x)$

Moment about B from force at A:

$M_a = 70(x+90)$

Moment from weight(which acts at the centre):

$M_w = 210(\frac{x+90}{2})$

The sum of these should be zero, and then you have an equation in x to solve!

Remember that anticlockwise moments are positive, and clockwise moments are negative!
Hi,

I dont understand why its $70(x+90)$ should it not be $70 \times xm$? because your taking moments about B, and the distance from B to A is 'x meters?

4. Oh sorry you edited your post, now it makes sense lol

I typed before you edited it

thanks a lot!

5. Originally Posted by Tweety
Oh sorry you edited your post, now it makes sense lol

I typed before you edited it

thanks a lot!
Sorry about the confusion. Slip of the fingers .

6. Am I doing this correctly?

here is my working for solving x

$140x + 70(x+90)= 210( \frac{x+90}{2})$

$210x + 6300= 210(x+45)$

$210x +6300 = (-210x) + (-9450)$????

this can't be right? what am i doing wrong?

7. Originally Posted by Tweety
Am I doing this correctly?

here is my working for solving x

$140x + 70(x+90)= 210( \frac{x+90}{2})$

$210x + 6300= 210(x+45)$

$210x +6300 = (-210x) + (-9450)$????

this can't be right? what am i doing wrong?
my answer book says the equation should be like this taking moments about A $140 \times 90 = 210 \times d$

d=60

8. The moments from the forces at A and C cause CLOCKWISE motion, which is taken as negative. The weight causes an ANTICLOCKWISE moment, which is positive.

Also you made an error when you went:

$210(\frac{90+x}{2}) = 210(45+x)$

Fraction do not work like that, it should be:

$210(\frac{90+x}{2}) = 210(45+\frac{x}{2})$

How's this look:

$-140(x)-70(x+90)+210(\frac{x+90}{2}) = 0$

$-140(x)-70x-70(90)+210(\frac{x}{2})+210(\frac{90}{2}) = 0$

$-140(x)-70x-70(90)+105x+105(90) = 0$

$-105x = 70(90)-105(90)$

$105x = 3150$

$x = 30cm$