A mass m1 = 6.6 kg rests on a frictionless table. It is connected by a massless

Question

A mass m1 = 6.6 kg rests on a frictionless table. It is connected by a massless and frictionless pulley to a second mass m2 = 3.7 kg that hangs freely. 1) What is the magnitude of the acceleration of block 1? 2) What is the tension in the string? 23.4 N Now the table is tilted at an angle of = 76° with respect to the vertical. Find the magnitude of the new acceleration of block 1. 2 m/s2 4) At what “critical” angle will the blocks NOT accelerate at all? 55.9 ° 5) Now the angle is decreased past the “critical” angle so the system accelerates in the opposite direction. If = 23° find the magnitude of the acceleration. The only part I can not figure out is 5

Explanation / Answer

Given,

m1 = 6.6 kg ; m2 = 3.7 kg

1)the acceleration will be:

a = m2g/(m1 + m2) = 3.7 x 9.8/(6.6 + 3.7) = 3.52 m/s^2

Hence, a = 3.52 m/s^2

2)T be the tension,

T = m2(g - a) = 3.7(9.81 - 3.52) = 23.3 N

Hence, T = 23.3 N

3)Now the net force is:

Fnet = m2g - m1g sin(theta)

Fnet = 3.7 x 9.8 - 6.6 x 9.8 x sin(90 - 76) = 20.61 N

a = Fnet/(m1 + m2) = 20.61/(6.6 + 3.7) = 2 m/s^2

Hence, a = 2 m/s^2

4)this condition occurs when

m1 g sin(theta) = m2g

sin(theta) = m2/m1 = 3.7/6.6 = 0.561

theta = sin^-1(0.561) = 34.1 deg

with ertical, theta = 90 - 34.1 = 55.9 deg

Hence, theta = 55.9 deg

5)theta = 23 deg

Fnet will be

Fnet = m1g sin(90 - theta) - m2g

Fnet = 6.6 x 9.8 x sin(90 - 23) - 3.7 x 9.8 = 23.28

a = Fnet/(m1 + m2) = 2.26 m/s^2

Hence, a = 2.26 m/s^2

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