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Differential Equations 18 Online
OpenStudy (anonymous):

solve the pink one

OpenStudy (anonymous):

OpenStudy (mathmate):

Blue: Denote tension in cord as T. Vertical reaction on bar equals weight of bar, assuming no friction anywhere. Take moments about B. Solve for T.

OpenStudy (anonymous):

how?

OpenStudy (anonymous):

can you show to me

OpenStudy (mathmate):

"How" meaning to take moments?

OpenStudy (anonymous):

yes but i'm confuse >.<

OpenStudy (anonymous):

roller gve vertical force to the wall so therefore

OpenStudy (anonymous):

|dw:1375615728179:dw|

OpenStudy (anonymous):

|dw:1375615816627:dw|

OpenStudy (anonymous):

also

OpenStudy (mathmate):

I can give an example: |dw:1375616654183:dw| A ladder 5 m long leans agains a smooth wall at B and stands on a rough floor at A at a distance of 3 m from the wall as shown above. The mass of the ladder is 10 kg. Since we don't know the friction on the floor, we can take moments about A so the friction does not come in the equation. Let reaction at B = R Mass at the middle (C) = m Take moments about A, since the ladder is in equilibrium, sum of moments = 0. -R*4 + mg*(3/2)=0 Note: moment = force * distance, clockwise is positive. Solve for R: R=(3mg/2)/4, or =3mg/8 N.

OpenStudy (mathmate):

|dw:1375617070605:dw|

OpenStudy (anonymous):

yes u are right

OpenStudy (mathmate):

So you're good for both problems?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

lets continue master

OpenStudy (anonymous):

|dw:1375616151580:dw|

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