Asked by Tsuki
A 100 kg box slides with constant velocity down a ramp inclined at 20 degrees, with the horizontal. Find the frictional force acting on the box.
Answers
Answered by
Henry
Wb = mg = 100kg * 8.8N/kg = 980 N. = Wt. of box.
Fb = 980N @ 20 Deg. = Force of the box.
Fp = 980*sin20 = 335.2 N. = Force parallel to ramp.
Fv = 980*cos20 = 920.9 N. = Force perpendicular to ramp.
Fk = Force of kinetic frictin.
Fn = Fp - Fk = ma
335.2 - Fk = 100*0.
335.2 - Fk = 0.
Fk = 335.2 N.
Fb = 980N @ 20 Deg. = Force of the box.
Fp = 980*sin20 = 335.2 N. = Force parallel to ramp.
Fv = 980*cos20 = 920.9 N. = Force perpendicular to ramp.
Fk = Force of kinetic frictin.
Fn = Fp - Fk = ma
335.2 - Fk = 100*0.
335.2 - Fk = 0.
Fk = 335.2 N.
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