In a real system of levers, wheels, or pulleys, the AMA is less than the IMA because _____.
2 answers:
In a real system of levers, wheel or pulleys, the AMA (actual mechanical advantage) is less than the IMA (ideal mechanical advantage) because of the presence of friction.
In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):
however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.
In fact, the IMA and the AMA of a machine are defined as the ratio between the output force (the load) and the input force (the effort):
however, the difference is that the IMA does not take into account the presence of frictions, while the AMA does. As a result, the output force in the AMA is less than the output force in the IMA (because some energy is dissipated due to friction), and the AMA is less than the IMA.
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Answer:
The final velocity of the object is, = 27 m/s
Explanation:
Given,
The acceleration of the object, a = 1000 m/s²
The initial displacement of the object, = 0 m
The final displacement of the object, = 0.75 m
The initial velocity of the object will be, = o m/s
The final velocity of the object, = ?
The average velocity of the object,
v = ( -
)/ t
= 0.75 / t
The acceleration is given by the relation
a = v / t
1000 m/s² = 0.75 / t²
t² = 7.5 x 10⁻⁴
t = 0.027 s
Using the I equation of motion,
= u + at
Substituting the values
= 0 + 1000 x 0.027
= 27 m/s
Hence, the final velocity of the object is, = 27 m/s
The required torque at the axle, is given by the difference between the
moments of the applied forces.
The torque required is <u>19.62 N·m counterclockwise</u>
Reasons:
The given parameters are;
Mass of the disk, m = 5.0 kg
Location of the axle = Half the radius of the disk
Diameter of the disk, D = 40 cm = 0.4 m
Applied mass, 0.1 m from the axle = 15 kg
Applied mass, 0.3 m from the axle = 10 kg
Required:
Magnitude of torque at the axle that prevent the disk from rotating
Solution:
Torque needed = Clockwise moment - Counterclockwise moment
Clockwise moment = (10 kg × 0.3 m + 5 kg × 0.1 m) × 9.81 m/s² = 34.335 N·m
Counterclockwise moment = 15 kg × 0.1 m × 9.81 m/s² = 14.715 N·m
τ + Counterclockwise moment = Clockwise moment
τ + 14.715 N·m = 34.335 N·m
Torque required, τ = 34.335 N·m - 14.715 N·m = 19.62 N·m
Torque required, τ = <u>19.62 N·m counterclockwise</u>
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<em>The probable question drawing obtained from a similar question online is attached</em>
