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October 21, 2014

October 21, 2014

Posted by **Danielle** on Sunday, March 11, 2012 at 3:40pm.

Any help with this questions, would be greatly appreciated. Thanks!

- Physics -
**Henry**, Tuesday, March 13, 2012 at 1:05amWs = mg = 145kg * 9.8N/kg = 1421 N. = Wt of sled incl. load.

Forward = 0 Deg.

Fs = 1421N @ 0 Deg. =Force of sled incl. load.

Fp=1421*sin(0)=0=Force parallel to gnd.

Fv = 1421*cos(0) = 1421 N. = Force perpendicular to gnd.

F1 = 83N @ 15.5 Deg.,CCW.

F2 = 75N @ 351 Deg.,CCW.

F3 = 77N @ 348 Deg.,CCW.

X=hor.=83*cos15.5+75*cos351+77*cos348=

229.4 N.

Y = 83*sin15.5+75*sin351+77*sin348=-5.56 N.

tanA = Y/X=(-5.56) / 229.4 = =0.024237

A = -1.39 Deg.,CW. = -1.39 + 369=358.6

Deg.,CCW.

Fap = X/cosA = 229.4 / cos358.6 = 229.5 N. @ 358.6 Deg. = Force applied.

Fn = Fap - Fp -Fk = ma = 0. a = 0.

229.5*cos358.6 - 0 - 1421u = 0.

1421u = 229.5*cos358.6 = 229.43.

u = 229.43 / 1421 = 0.161 = Coefficient of kinetic friction.

- Physics -
**Danielle**, Tuesday, March 13, 2012 at 4:21pmThank you so much for your help! But just one question what does CCW mean? And once again, I appreciate your help immensly!

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