2. Block 1 is placed on a 30 degree incline, and is attached to block 2 by a rop
ID: 1394822 • Letter: 2
Question
2. Block 1 is placed on a 30 degree 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 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 mus = 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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