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 θ = 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 θ = 23° find the magnitude of the acceleration. The only part I can not figure out is 5
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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