For this problem, use g = 10 N/kg. As shown in the figure above, a pulley is mou

Question

For this problem, use g = 10 N/kg. As shown in the figure above, a pulley is mounted to the ceiling of an elevator is initially traveling with a constant velocity of 1.90 m/s, directed up. The rope (which we assume to have no mass) passing over the pulley has block A tied to one end block B tied to the other. The mass of block A is 4.60 kg. Block A remains on the floor of the elevator, at rest with respect to the elevator. The mass of block B is 1.70 kg, and block B is also at rest with respect to the elevator, hanging from the rope. Calculate the magnitude of the tension in the rope. Calculate the magnitude of the force exerted on block A by the floor of the elevator.

Explanation / Answer

here,

case a:
as block A and blcok B are in rest therefore
T = Mag + Mbg
T = 4.6 * 10 + 1.70 * 10
T = 63 N
Tension in rope is 63N

Case B
Normal force = mass * acceleration due to gravity
N = 4.6 * 10 = 46 N
Force exereted by elevator on blockA is 46N

Case C :
Magnitude of force exterted by ceiling on pully systemis
N = Mp*g = 0.8 *10 = 8 N

Magnitude of force exerted by ceiling of elevator onpulley support is 8N

Case D :
When elevator start accelerating upward with 1.60 m/s^2

Then total W = (g+a)(Ma+Mb)
Net force = 0
T = (g+a)(Ma+Mb)
T = (10+1.60)(4.6+1.70)
T = 73.08 N

The magnitude in rope when elevator accelerating upward is 73.08N

case E:

N(force exerted by floor on Block A) = Ma(g+a)
N = 4.6(10+1.6)
N = 53.36 N

Magnitude of force exerted by floor on Block A is 53.36N

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