Asked by amy
A boy drags a 100 N sled up a 20 degree slope at constant velocity. If the coefficient of friction between the sled and hill is 0.2 what force must he exert at an angle of 35 degrees with respect to the hill?
Answers
Answered by
Henry
Fs = 100N @ 20 deg.
Fp = 100sin20 = 34.2N. = Force parallel
to the hill.
Fv = 100cos20 = 94N = Force perpendicular to the hill.
Ff = 0.2*94 = 18.8N = Force of friction.
Fap*cos35 - Fp - Ff = 0,
Fap*cos35 - 34.2 - 18.8 = 0,
Fap*cos35 - 53 = 0,
Fap*cos35 = 53,
Fap = 64.7N = Applied force.
Fp = 100sin20 = 34.2N. = Force parallel
to the hill.
Fv = 100cos20 = 94N = Force perpendicular to the hill.
Ff = 0.2*94 = 18.8N = Force of friction.
Fap*cos35 - Fp - Ff = 0,
Fap*cos35 - 34.2 - 18.8 = 0,
Fap*cos35 - 53 = 0,
Fap*cos35 = 53,
Fap = 64.7N = Applied force.
Answered by
Eric
Henry, you didn't account for the vertical pull of the boy. therefore friction is less than 18.8.
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