Posted by Millie on Friday, November 18, 2011 at 4:24pm.
A 26.0 kg block is connected to an empty 1.00 kg bucket by a cord running over a frictionless pulley. The coefficient of static friction between the table and the block is 0.490 and the coefficient of kinetic friction between the table and the block is 0.320. Sand is gradually added to the bucket until the system just begins to move.
Part A of the question asked me to calculate the mass of sand added to the bucket which i determined to be 11.7 kg.
Part B asks the acceleration of the system downward. Can someone help?
- Physics - Henry, Sunday, November 20, 2011 at 9:42pm
Wb = mg = 26kg * 9.8N/kg = 254.8N.
Fb = 254.8N @ 0deg.,
Fp = 254.8sin(0) = 0 = Force parallel to table.
Fv = 254.8cos(0) = 254.8N. = Force perpendicular to table.
Fs = u*Fv = 0.490 * 254.8 = 124.9N. =
Force of static friction.
Fk = 0.320 * 254.8 = 81.5N. = Force of
A. Fap - Fp - Fs = 0,
Fap - 0 - 124.9 = 0,
Fap = 124.9N. = Force applied = Force of sand plus bucket.
mg = Fap,
m = Fap/g = 124.9/9.8 = 12.7kg = mass of bucket plus sand.
Ms = 12.7 - 1 = 11.7kg = Mass of sand.
B. Fn = Fap - Fp - Fk,
Fn = 124.9 - 0 - 81.5 = 43.4N. = Net
a = Fn / (m1+m2) = 43.4 / (26+12.7) =
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