Hey Everyone, I worked out this problem for a class and got it incorrect, I just need to know what I got it wrong?
A 27.0 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 0.435 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.
(a) Calculate the mass of sand added to the bucket.
Answer in kg
(b) Calculate the acceleration of the system.
Answer in m/s2
Max Friction at rest = .435 * 27 = 11.745
Tension in String = mass of bucket+ mass of sand) * g
= 11.745/9.81 = 1.1972
mass of sand = 1.1972-1= .1972kg
Once moving... 0.32*27=8.64
11.745-8.64=3.105
Force =mass*acceleration
acceleration = f/mass = 3.105/27
acc= .115 ms2
Can someone please help me figure out what I did incorrectly. :(
A 27.0 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 0.435 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.
(a) Calculate the mass of sand added to the bucket.
Answer in kg
(b) Calculate the acceleration of the system.
Answer in m/s2
Max Friction at rest = .435 * 27 = 11.745
Tension in String = mass of bucket+ mass of sand) * g
= 11.745/9.81 = 1.1972
mass of sand = 1.1972-1= .1972kg
Once moving... 0.32*27=8.64
11.745-8.64=3.105
Force =mass*acceleration
acceleration = f/mass = 3.105/27
acc= .115 ms2
Can someone please help me figure out what I did incorrectly. :(
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a)
Force needed to overcome static friction
= 0.435 * 27 = 11.745 kg
Tension in string = 11.745 kg
Weight of sand = 11.745 - 1 = 10.745 kg
b)
Total mass being accelerated
= 27 + 11.745 = 38.745 kg
Force required to overcome kinetic friction
= 0.320 * 27 = 8.64 kg
Effective force accelerating system
= 11.745 - 8.64 = 3.105 kg
Acceleration = 9.81 * 3.105 / 38.745
= 0.7862 m/s^2 (to 4 decimal places)
Force needed to overcome static friction
= 0.435 * 27 = 11.745 kg
Tension in string = 11.745 kg
Weight of sand = 11.745 - 1 = 10.745 kg
b)
Total mass being accelerated
= 27 + 11.745 = 38.745 kg
Force required to overcome kinetic friction
= 0.320 * 27 = 8.64 kg
Effective force accelerating system
= 11.745 - 8.64 = 3.105 kg
Acceleration = 9.81 * 3.105 / 38.745
= 0.7862 m/s^2 (to 4 decimal places)
-
Your statement:
Max Friction at rest = .435 * 27 = 11.745
is wrong.
The block's weight = mg = 27.0x9.81 = 264.87N
Normal force on block, N = weight = 264.87N
Limiting frictional force = μs.N so you should have said:
Max Friction at rest = 0.435 * 264.87 = 115.2N
Your method for the rest of part a) is correct, but what you have written down is muddled. You should have written something like this:
Tension in string = (mass of bucket+ mass of sand) * g
115.2 = (1 + m) x 9.81
You should then solve for m.
___________________________
b) The method is wrong. As well as incorrectly calculating the kinetic friction force (same mistake as part a), you have assumed the tension is the same as in part a). When the system is accelerating, the tension changes. This is how you find the tension (T) and the acceleration (a):
Kinetic friction = μs.N = 0.32 * 264.87 = 84.76N
Resultant horizontal force on block = T - 84.76
Applying F = ma to block
T - 84.76 = 27a (equation 1)
Resultant vertical force on bucket + sand = (1+m)g - T (where m is the value from part a)
Applying F = ma to bucket + sand
(1+m)g - T = (1+m)a (equation 2)
To find a, solve the simultaneous equations, 1 and 2.
Max Friction at rest = .435 * 27 = 11.745
is wrong.
The block's weight = mg = 27.0x9.81 = 264.87N
Normal force on block, N = weight = 264.87N
Limiting frictional force = μs.N so you should have said:
Max Friction at rest = 0.435 * 264.87 = 115.2N
Your method for the rest of part a) is correct, but what you have written down is muddled. You should have written something like this:
Tension in string = (mass of bucket+ mass of sand) * g
115.2 = (1 + m) x 9.81
You should then solve for m.
___________________________
b) The method is wrong. As well as incorrectly calculating the kinetic friction force (same mistake as part a), you have assumed the tension is the same as in part a). When the system is accelerating, the tension changes. This is how you find the tension (T) and the acceleration (a):
Kinetic friction = μs.N = 0.32 * 264.87 = 84.76N
Resultant horizontal force on block = T - 84.76
Applying F = ma to block
T - 84.76 = 27a (equation 1)
Resultant vertical force on bucket + sand = (1+m)g - T (where m is the value from part a)
Applying F = ma to bucket + sand
(1+m)g - T = (1+m)a (equation 2)
To find a, solve the simultaneous equations, 1 and 2.