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

[cracktheegg]

Homework Statement

[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]

Homework Equations

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

M=F*D

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

 

[adjacent]

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

 

[cracktheegg]

sry, forget abt the question

 

[cracktheegg]

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

 

[SteamKing]

ƩM$_{A}$ = Ʃ r x F

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

 

[cracktheegg]

is 20N moment = -20(6)?

 

[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?

 

[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.

20N: 20*6 =60NM

Check the arithmetic.

30N: -97.53Nm

Please show how you got to that number. Why negative?

40N: -40sin45 *2.83=80.04NM

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.

10N: 10sin15(6)=15.6NM

Sign?

What about the two Fs?

 

[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

 

[haruspex]

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

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?

 

[cracktheegg]

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

 

[adjacent]

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

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

 

[cracktheegg]

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

correct?

 

[haruspex]

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

correct?

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

 

[cracktheegg]

ty guys, you guys help me alot