Posted by Danielle on Sunday, March 11, 2012 at 3:40pm.
Three huskies are pulling a dog sled. The mass of the driver, sled and supplies is 145 kg. Snowy is pulling with a force of 83 N and 15.5degrees to the left of forward. Buster is pulling with a force of 75 N at 9degrees right of forward and Prince is pulling with a force of 77N at 12 degrees right of forward. The sled is moving forward at a constant velocity. What is the coefficient of kinetic friction between the sled and the snow?
Any help with this questions, would be greatly appreciated. Thanks!

Physics  Henry, Tuesday, March 13, 2012 at 1:05am
Ws = 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 4:21pm
Thank 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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