The suspended 2.9 kg mass on the right is moving up, the 1.9 kg mass slides down

Question

The suspended 2.9 kg mass on the right is moving up, the 1.9 kg mass slides down the ramp, and the suspended 8.1 kg mass on the left is moving down. There is friction between the block and the ramp. The acceleration of gravity is 9.8 m/s2 . The pulleys are massless and frictionless. What is the tension in the cord connected to the 8.1 kg block? Answer in units of N.

A person stands on a scale in an elevator. The maximum and minimum scale readings are 827.9 N and 328.8 N, respectively. The acceleration of gravity is 9.8 m/s2 . Assume the magnitude of the acceleration is the same during starting and stopping, and determine the acceleration of the elevator. Answer in units of m/s2.

Explanation / Answer

given M = 2.9 kg, on the right moving up

m = 1.9 kg, slides down the ramp

m' = 8.1 kg, moving down to the left

tension in chord with the 8.1 kg mass = T'

tension in the chord connecte to the 2.9 kg mass = T

hence

from force balance

m'g - T' = m'a

T' + mg*sin(theta) - T = ma

T - Mg = Ma

hence

mg*sin(theta) - Mg + m'*g = a(m + m' + M)

hence

a = g(m*sin(theta) - M + m')/(m + m' + M)

now the value of angle theta is not avaible and the coefficient of friciton is not given as well so the answer cannot be numerically found

2. Wmax = 827.9 N

Wmin = 328.8 N

g = 9.8 m/s/s

acceleration while stopping and starting is the same ( n magnitude) = a

hence

Wmax = m(g + a)

Wmin = m(g - a)

hence

(Wmax - Wmin) = m(2a)

(Wmax + Wmin) = 2mg

hence

(Wmax - Wmin)/(Wmax + Wmin) = a/g

hence

a = 4.232878 m/s/s

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