# How to solve the moment questions, I tried couple of times but

1. Nov 21, 2013

### cracktheegg

1. The problem statement, all variables and given/known data

[URL=http://s1345.photobucket.com/user/Duk_Bato/media/Untitled_zps863917c8.png.html][PLAIN]http://i1345.photobucket.com/albums/p679/Duk_Bato/Untitled_zps863917c8.png[/URL][/PLAIN]

2. Relevant equations

Tan angle = opposite/adjacent Cos angle= adjacent/hypotenuse
Sin angle= opposite/hypotenuse

M=F*D

3. The attempt at a solution

40(2.83 sin45) -30(4.47 cos26.57) -20cos45 * 0.6 +10*0.6 sin15
=-46.82

which is not the correct answer

2. Nov 21, 2013

### adjacent

This is too vague.What the question?
Moment is not just F*D
Moment is force x perpendicular distance.

3. Nov 21, 2013

### cracktheegg

sry, forget abt the question

4. Nov 21, 2013

### cracktheegg

Find the moment about point hinge A due to all known forces

5. Nov 21, 2013

### SteamKing

Staff Emeritus
ƩM$_{A}$ = Ʃ r x F

where r is the position vector from the force to A and F is the force vector.

6. Nov 21, 2013

### cracktheegg

is 20N moment = -20(6)?

7. Nov 21, 2013

### cracktheegg

20N =20*6 =60NM
30N= -97.53Nm
40N= -40sin45 *2.83=80.04NM
10N = 10sin15(6)=15.6NM

Can anyone tell me any of the forces are wrong?

8. Nov 21, 2013

### haruspex

It's a little confusing to use equals signs like that... edited.
I assume anticlockwise is positive in your scheme.
Note the distances in the diagram are cm, not m.
Check the arithmetic.
Please show how you got to that number. Why negative?
By plugging in numbers too soon you introduced a rounding error. It should be exactly 80. Can you see an easy way to obtain that? And you lost the minus sign.
Sign?

What about the two Fs?

9. Nov 21, 2013

### cracktheegg

20N has a moment of 20*0.06 = +1.2Nm

x component of force moment is -30cos(60° )*0.02

y component of force moment is +30sin(60° )*0.04

40N has a moment -40*0.02 = -80Nm

10N has a moment of -10sin(15°)*6

10. Nov 21, 2013

### haruspex

That all looks right. I still don't understand where the two 'F' forces come in. Do they arise in a different part of the question?

11. Nov 21, 2013

### cracktheegg

There is question B, which ask what magnitude of the F will keep it equilibrium.

12. Nov 22, 2013

### adjacent

If it is in equilibrium,Anticlockwise moment should be equal to clockwise moment.

13. Nov 22, 2013

### cracktheegg

91.96= Fsin60 5.67-Fcos15
F= 91.96/ (sin60 5.67 -cos15)
=27.4N

correct?

14. Nov 22, 2013

### haruspex

I get 27.48, but that's probably near enough.

15. Nov 22, 2013

### cracktheegg

ty guys, you guys help me alot

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