The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. Assume that the rope and pu
lley are massless and that there is no friction in the pulley. If the masses have the values m1=21.1 kg and m2=12.9 kg, find the magnitude of their acceleration ???? and the tension T in the rope. Use ????=9.81 m/s2.
1 answer:
Answer:
T=157.06 N
Explanation:
Given that
Mass of first block = 21.1 kg
Mass of second block = 12.9 kg
First mass is heavier than first that is why mass second first will go downward and mass second will go upward.
Given that pulley and string is mass less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.
So now from force equation
21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a
Lets tension in string is T
T=21.1(9.81-2.36) N
T=157.06 N
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Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
