Posted by **Jennifer** on Wednesday, December 8, 2010 at 12:19am.

A 10N meterstick is suspended by two sping scales, one at the .08 m mark and the other at the .9 m mark. If a weight of 5N is hung at the .2 m mark and a weight of 17N is hung at the .55 m mark, what will be the reading on each scale?

I have:

F(_1)+F(_2)=5+10+17=32

With moment at F(_1), sum of torques=0:

F(_2)(.82)-5(.2)-10(.5)-17(.55)=0

.82F(_2)-1-5-9.35=0

.82F(_2)=15.35

F(_2)=18.72

This makes F(_1)=13.28

Do I now use F=ma as follows to find the scale reading?

F(_1)=m(_1)a

13.28=m(_1)(9.8)

m(_1)=1.355 or about 1.4 kg

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