Asked by Lian
A 1200 lb sleigh is pulled along a horizontal surface at uniform speed by means of a rope that makes an angle of 30 degrees above the horizontal. If the tension in the rope is 100 lb., what is the coefficient of friction?
Answers
Answered by
Henry
Ws = 1200 Lbs.
Fp = 100*Cos30 = 86.6 Lbs. = Force parallel with the surface.
Fn = 1200-100*sin30 = 1150 Lbs. = Normal force.
Fk = u*Fn = u*1150 Lbs.
Fp-u*Fn = Ws*a.
u*Fn = Fp-Ws*0,
u*Fn = Fp, u = Fp/Fn = 86.6/1150 = 0.075.
Fp = 100*Cos30 = 86.6 Lbs. = Force parallel with the surface.
Fn = 1200-100*sin30 = 1150 Lbs. = Normal force.
Fk = u*Fn = u*1150 Lbs.
Fp-u*Fn = Ws*a.
u*Fn = Fp-Ws*0,
u*Fn = Fp, u = Fp/Fn = 86.6/1150 = 0.075.
Answered by
may
0.075
Answered by
molina
A ladder has a weight of 30-lbs and a length of 26-ft. Determine the maximum distance D it can be
placed from the smooth wall and not slip. The coefficient of static friction between the floor and the
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placed from the smooth wall and not slip. The coefficient of static friction between the floor and the
pole is 0.3.
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