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1. Homework Statement
Blocks A and B are connected by a cable that passes over support C. Friction between the blocks and the inclined surfaces may be neglected. Knowing that motion of block B up the incline is impending when m_{B} = 4.5 kg, determine:
(a) the coefficient of static friction between the rope and the support
(b) the largest value of m_{B} for which equilibrium is maintained
2. Homework Equations
T_{2} = T_{1}e^(μ_{s}β) (1)
β is in angle in radians
W = mg
3. The Attempt at a Solution
My problem is that for part (b), my answer is wrong. I would like to know where I went wrong in my calculations. Thanks for the help in advance.
I found the forces working on Block A
ƩF_{x} = 0
→ T_{A}  W_{A}sin(16) = 0
T_{A} = 24.336 N
Then, Block B
ƩF_{x} = 0
→ W_{B}sin(16)  T_{B} = 0
T_{B} = 12.168 N
Then, I looked at support C where I found the angle to be 32°, and I used (1) to find the static friction.
β = (32)(∏/180) ≈ 0.559
T_{A} = T_{B}e^(μ_{s}β)
24.336 N = (12.168 N)e^(μ_{s}β)
e^(μ_{s}β) = (24.336 N)/(12.168 N)
μ_{s}β = ln [(24.336 N)/(12.168 N)]
μ_{s} = (1/0.559)*ln [(24.336 N)/(12.168 N)]
μ_{s} ≈ 1.240
Then, since I know T_{B} has to be greater than itself to upset the equilibrium, I did
T_{B} = T_{A}e^(μ_{s}β)
T_{B} = (24.336 N)e^[(1.240)(0.559)]
T_{B} ≈ 48.673 N
Then, to get the mass I divided the T_{B} by g = 9.81
m_{B} = (48.673)/(9.81) ≈ 4.962 kg
That is wrong.
Blocks A and B are connected by a cable that passes over support C. Friction between the blocks and the inclined surfaces may be neglected. Knowing that motion of block B up the incline is impending when m_{B} = 4.5 kg, determine:
(a) the coefficient of static friction between the rope and the support
(b) the largest value of m_{B} for which equilibrium is maintained
2. Homework Equations
T_{2} = T_{1}e^(μ_{s}β) (1)
β is in angle in radians
W = mg
3. The Attempt at a Solution
My problem is that for part (b), my answer is wrong. I would like to know where I went wrong in my calculations. Thanks for the help in advance.
I found the forces working on Block A
ƩF_{x} = 0
→ T_{A}  W_{A}sin(16) = 0
T_{A} = 24.336 N
Then, Block B
ƩF_{x} = 0
→ W_{B}sin(16)  T_{B} = 0
T_{B} = 12.168 N
Then, I looked at support C where I found the angle to be 32°, and I used (1) to find the static friction.
β = (32)(∏/180) ≈ 0.559
T_{A} = T_{B}e^(μ_{s}β)
24.336 N = (12.168 N)e^(μ_{s}β)
e^(μ_{s}β) = (24.336 N)/(12.168 N)
μ_{s}β = ln [(24.336 N)/(12.168 N)]
μ_{s} = (1/0.559)*ln [(24.336 N)/(12.168 N)]
μ_{s} ≈ 1.240
Then, since I know T_{B} has to be greater than itself to upset the equilibrium, I did
T_{B} = T_{A}e^(μ_{s}β)
T_{B} = (24.336 N)e^[(1.240)(0.559)]
T_{B} ≈ 48.673 N
Then, to get the mass I divided the T_{B} by g = 9.81
m_{B} = (48.673)/(9.81) ≈ 4.962 kg
That is wrong.
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