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Statics - Moments

by ZaZu
Tags: moments, statics
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ZaZu
#1
Mar2-09, 05:02 PM
P: 36
1. The problem statement, all variables and given/known data

I got a problem solving this question.

http://img228.imageshack.us/img228/7049/image164.jpg


3. The attempt at a solution

Im not sure if im missing a rule here. I take CW to be positive and CCW to be negative.
But my attempt to solve it failed =[

http://img220.imageshack.us/img220/7638/image165.jpg


Your help is really appreciated :)
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PhanthomJay
#2
Mar2-09, 05:10 PM
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Looks like you forgot to multiply by the moment arm perpendicular distance to the pivot. Also, check your signs for each of the torques.
ZaZu
#3
Mar2-09, 05:11 PM
P: 36
Quote Quote by PhanthomJay View Post
Looks like you forgot to multiply by the moment arm perpendicular distance to the pivot. Also, check your signs for each of the torques.
Yes, the distance x, correct ?? Its M= F . x .. But how can I find x ? :S

PhanthomJay
#4
Mar2-09, 05:17 PM
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Statics - Moments

A torque of a force or force components is calculated by multiplying the force or force components times the perpendicular distance between the line of action of that force and the pivot. You alreadty I think have the vertical component of each force. So what's the perpendicular distance, and what are the signs of the torques?
ZaZu
#5
Mar2-09, 05:20 PM
P: 36
The first force is acting at a clockwise direction, it is positive.
The second force is acting at an anti-clockwise direction, it is negative.

Regarding the Perp. distance, its just not clicking in my head.
ZaZu
#6
Mar2-09, 05:26 PM
P: 36
Ok wait, do you mean by perp. distance is the distance given between the pivot and the force itself ?? Which is 3x10^-3 for F1 and 3.8x10^-3 for F2 ?
PhanthomJay
#7
Mar2-09, 05:28 PM
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No, you are looking for the torque of both forces about the NUT. They both seem ccw, no? Also, regarding perpendicular distance from the horizontal force components to the nut, they are ___ and ___ for each horizontal force component?
ZaZu
#8
Mar2-09, 05:33 PM
P: 36
Ok I think i got it :D :D

http://img80.imageshack.us/img80/2008/image167k.jpg

Im not sure about the final result though, is it N or Kn or Nm ..
The given answer is 81.1N .. I found 8116 .. hmm if the supposed answer is 81.1Kn then my answer would be correct for N.

Correct me if im mistaken.
PhanthomJay
#9
Mar2-09, 06:03 PM
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Quote Quote by ZaZu View Post
Ok I think i got it :D :D

http://img80.imageshack.us/img80/2008/image167k.jpg

Im not sure about the final result though, is it N or Kn or Nm ..
The given answer is 81.1N .. I found 8116 .. hmm if the supposed answer is 81.1Kn then my answer would be correct for N.

Correct me if im mistaken.
Watch your units. If the torque is in N.m, then Force should be in Newtons and distance should be expressed in meters.
ZaZu
#10
Mar2-09, 06:26 PM
P: 36
Quote Quote by PhanthomJay View Post
Watch your units. If the torque is in N.m, then Force should be in Newtons and distance should be expressed in meters.
So where does that leave me ?? Am I supposed to convert or anything ?? If I divide by 100 I get the correct answer. But I want to know why am I dividing by a hundred.

Can you please tell me ? I got test tomorrow and its getting late =\
I really need to know how to correct my mistakes !
ZaZu
#11
Mar3-09, 02:51 AM
P: 36
Bump =\
PhanthomJay
#12
Mar3-09, 07:42 AM
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Quote Quote by ZaZu View Post
So where does that leave me ?? Am I supposed to convert or anything ?? If I divide by 100 I get the correct answer. But I want to know why am I dividing by a hundred.

Can you please tell me ? I got test tomorrow and its getting late =\
I really need to know how to correct my mistakes !
It must be your algebra:
Fsin60(.3m) + Fsin70(.38m) = 50N.m
F= 80.1N


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