Two blocks are connected over a pulley. The mass of block A is 10 kg, and the coefficient of kinetic friction between A and the incline is 0.20. Angle θ is 30°. Block A slides down the incline at constant speed. What is the mass of block B?
-
Since A is moving down the incline,
the force of friction is uphill with magnitude
fk = μkFn
Since the block moves with constant speed, a = 0.
According to Newton's Second Law,
ΣFx = T + fk - mBg sinθ = 0 (1)
ΣFy = Fn - mBg cosθ = 0 (2)
mAg - T = 0 (3)
Solving (1) for T,
T = mBg sinθ - fk
or
T = mBg sinθ - μkmBg cosθ
Solving (3) for mA,
mA = T/g
Substituting for T,
mA = (mBg sinθ - μkmBg cosθ) / g
or
mA = mB sinθ - μkmB cosθ
or
mA = mB(sinθ - μk cosθ)
mA = (10 kg)[sin 30 - (0.2)cos 30]
mA = 3.3 kg
the force of friction is uphill with magnitude
fk = μkFn
Since the block moves with constant speed, a = 0.
According to Newton's Second Law,
ΣFx = T + fk - mBg sinθ = 0 (1)
ΣFy = Fn - mBg cosθ = 0 (2)
mAg - T = 0 (3)
Solving (1) for T,
T = mBg sinθ - fk
or
T = mBg sinθ - μkmBg cosθ
Solving (3) for mA,
mA = T/g
Substituting for T,
mA = (mBg sinθ - μkmBg cosθ) / g
or
mA = mB sinθ - μkmB cosθ
or
mA = mB(sinθ - μk cosθ)
mA = (10 kg)[sin 30 - (0.2)cos 30]
mA = 3.3 kg