A block slides down a frictionless incline making an angle theta with floor at an elevator.The elevator is descending with acceleration a.The value of normal reaction acting on block is?
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Answers
Answer:
Let the acceleration of the block kept on an incline is b and its mass is m.
Since the lift is retarding downwards with a magnitude a, a pseudo force will be applied on the block in downward direction as shown in the fbd.
(ma+mg)sinθ=mb
b=(a+g)sinθ
solution
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Answer:
The normal reaction acting on block is N=m(g-a)cosФ
where g>a
Explanation:
When the block slides on a horizontal frictionless route no paintings is finished with the aid of using the gravitational pressure and no extrade happens for the kinetic strength or ability strength of the block.
The ordinary pressure may be observed the use of the gravitational, or weight, pressure and the attitude of the willing plane. It is given with the aid of using F=mgcosθ, wherein m is the mass, g is acceleration because of gravity and θ (theta) is the attitude.
The coefficient of friction among the frame and the floor is µs. Let the preliminary cost of θ be 0 and if we slowly begin growing the cost of θ, then at a specific cost of θ = φ the block simply begins offevolved to move. This cost of θ =φ is known as the attitude of friction.
for the reason that frame is sliding down at an attitude Ф,
N+ ma cosФ = mg cosФ
mg sin Ф > ma sin Ф
N=m(g-a)cosФ
wherein g>a
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