2. Block 1 is placed on a 30 incline, and is attached to block 2 by a rope. Bloc
ID: 1394824 • Letter: 2
Question
2. Block 1 is placed on a 30 incline, and is attached to block 2 by a rope. Block 2 is hanging from a pulley over the top edge of the incline and is also completely submerged in a tank of water. a. (5 pts.) For what block 2 density could 5 kg static equilibrium be maintained without static friction? b. (5 pts.) If block 2 were aluminum, what is the smallest coefficient of static friction required to for static equilibrium? c. (5 pts.) If the coefficient of static friction were us 0.30, for what maximum and minimum block 2 densities could static equilibrium be maintained?Explanation / Answer
part (a)
for M1
T - M1*g*sin30 = M1*a
for M2
T + Fb - M2*g = M2*a
in static equlibrium
a = 0
therefore
T = M1*g*sin30
Fb = M2*g - M1*g*sin30
buoyancy force Fb = d_water*V*g
d_water*Volume = M2 - M1*sin30
1000*volume = 5 - 2
Volume = 0.003 m^3
density = M2/V = 5/0.003 = 1666.7 kg / m^3 <----------answer
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dAl = 2700 kg /m^3
for M1
T - us*M1*g*sin30 - M1*g*sin30 = Fnet
for M2
T + Fb - M2*g = Fnet
T - us*M1*g*cos30 - M1*g*sin30 = T + Fb - M2*g
us*M1*g*sin30 = M2*g - Fb - M1*g*sin30
Fb = dw*V*g
us*M1*g*cos30 = M2*g - dw*V*g - M1*g*sin30
us*M1*cos30 = dAl*V - dw*V - M1*sin30
us*4*cos30 = (2700*0.003)-(1000*0.003)-(4*sin30)
us = 0.89 <----------answer
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partc
M2 = d*V
for smallest
M1*g*sin30 - us*M1*g*cos30 = M2*g - Fb
M1*g*sin30 - us*M1*g*cos30 = d*V*g - dw*V*g
M1*sin30 - us*M1*cos30 = d*V - dw*V
(4*sin30)-(0.3*4*cos30) = (d*0.003)-(1000*0.003)
d = 1320 kg /m^3 <----------answer
for maximum
M1*g*sin30 + us*M1*g*cos30 = M2*g - Fb
M1*g*sin30 + us*M1*g*cos30 = d*V*g - dw*V*g
M1*sin30 + us*M1*cos30 = d*V - dw*V
(4*sin30)+(0.3*4*cos30) = (d*0.003)-(1000*0.003)
d = 2017.1 kg /m^3 <----------answer
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