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The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. The pulley can be approximated by a uniform disk with mass p=7.95 kg and radius p=0.89 m. The hanging masses are L=32.0 kg and R=17.8 kg. Calculate the magnitude of the masses' acceleration and the tension in the left and right ends of the rope, L and R , respectively.

Respuesta :

Answer:

Acceleration(a) = 2.588 m/s²

TL = 230.784 N

TR = 220.5 N

Explanation:

Given:

M = 7.95 kg

mL = 32 kg

mR = 17.8 kg

g = 9.8 m/s²

Find:

Acceleration(a)

TL

TR

Computation:

Acceleration(a) = [(mL - mR)g] / [mL + mR + M/2]

Acceleration(a) = [(32 - 17.8)9.8] / [32 + 17.8 + 7.95/2]

Acceleration(a) = [139.16] / [53.775]

Acceleration(a) = 2.588 m/s²

TL = mL(g-a)

TL = 32(9.8-2.588)

TL = 230.784 N

TR = mR(g+a)

TR = 17.8(9.8+2.588)

TR = 220.5 N

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