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.
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
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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