For this problem, use g-10 N/kg As shown in the figure above, a pulley is mounte

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. The elevator is initially traveling with a constant velocity of 1.80 m/s, directed up. The rope (which we assume to have no mass) passing over the pulley has block A tied to one end and 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 1.40 kg, and block B is also at rest with respect to the elevator, hanging from the rope. (a) Calculate the magnitude of the tension in the rope 14 (b) Calculate the magnitude of the force exerted on block A by the floor of the elevator 32 c) The total mass of the pulley and its support is 0.700 kg. Calculate the magnitude of the force exerted by the ceiling of the elevator on the pulley support. 34.4 At some later time, the elevator is moving, and it has an acceleration of 1.90 m/s2, directed down. The blocks remain at rest with respect to the elevator. (d) Calculate the magnitude of the tension in the rope now 1.77647XN (e) Calculate the magnitude of the force exerted on block A by the floor of the elevator now. 26.9176 X N

Explanation / Answer

(a)

Let, tension in the rope = T

By balancing the forces in verticle,

T = mB * g

T = 1.40 * 9.81

T = 14 N

(b)

Let,  force exerted by floor = F

By balancing force in vertical,

F + T = mA * g

F = 4.60*9.8 - 14

F = 32 N

(c)

total force exerted by the ceiling,

Fp = Wp + 2T

Fp = 0.70*9.8 + 2*14

Fp = 34.8 N

(d)

Let, tension in the rope now = T

Balance forces for block B,

WB - T = mB*a

1.40*9.8 - T = 1.40*1.90

T = 11.06 N

(e)

Balance forces for block A,

Let. force exerted on block A by the floor now = F

WA - T - F = mA * a

4.6*9.8 - 11.06 - F = 4.60*1.90

F = 25.3 N

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