Asked by anna
Suppose a 46-N sled is resting on packed snow. The coefficient of kinetic friction is 0.11. If a person weighing 625 N sits on the sled, what force is needed to pull the sled across the snow at constant speed?
Answers
Answered by
Henry
Ws = 46N. = Weight of sled.
Fs = 46N @ 0deg. = Force of sled.
Fp = 46sin(0) = 0=Force parallel to the snow.
Fv = 46cos(0) = 46N. = Force perpendicular to the snow.
Ff = u(Fv+625),
Ff = 0.11(46 + 625) = 671N. = Force of
friction.
Fn = Fap - Fp - Ff = 0,
Fap - 0 - 671 = 0,
Fap = 671N. = Force applied to pull sled.
Fs = 46N @ 0deg. = Force of sled.
Fp = 46sin(0) = 0=Force parallel to the snow.
Fv = 46cos(0) = 46N. = Force perpendicular to the snow.
Ff = u(Fv+625),
Ff = 0.11(46 + 625) = 671N. = Force of
friction.
Fn = Fap - Fp - Ff = 0,
Fap - 0 - 671 = 0,
Fap = 671N. = Force applied to pull sled.
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