- 05 Jun '06 00:08

Wouldn't be 80 X 5600 by any chance?*Originally posted by Knight Square***This is a FORCES question,**

f1=5600N

d1=100cm

f2= ??X??

d2=8000cm

Work out ??X?? and while your at it work out Turning MOMENT!

Hint: Moment = Force * Distance

ie. M=FD

Hint: Think of a see-saw with those numbers, numbers being f1,d1... - 05 Jun '06 00:42f2 = 7N

Also the turning moment provided by f2 is 7*80 = 560Nm.

(Note that you work in metres as helps prevent stupid errors).

This is of course assuming the system is in equilibrium (both rotationally due to the two moments from f1 and f2, and linearly due to a reaction force at the pivot). - 05 Jun '06 00:50

I was going to ask OP if xyr Teacher said it was in equilibrium and whether xyr teacher agreed.*Originally posted by XanthosNZ***f2 = 7N**

Also the turning moment provided by f2 is 7*80 = 560Nm.

(Note that you work in metres as helps prevent stupid errors).

This is of course assuming the system is in equilibrium (both rotationally due to the two moments from f1 and f2, and linearly due to a reaction force at the pivot). - 05 Jun '06 05:47

I know the question was not written out well.*Originally posted by Knight Square***This is a FORCES question,**

f1=5600N

d1=100cm

f2= ??X??

d2=8000cm

Work out ??X?? and while your at it work out Turning MOMENT!

Hint: Moment = Force * Distance

ie. M=FD

Hint: Think of a see-saw with those numbers, numbers being f1,d1...

The Answers:

f2=70N

MOMENT was equal in both direction

Moment = f1 X d1

or

Moment = f2 X d2

The answer being:

560000Ncm = 5600 X 100,

' ' = 8000 X 70,

The Answer should be:

5600Nm, because someone said that the measurements should be in metres.

Thanks