left of the pulley N right of the pulley N A block of mass m_1 = 2.05 kg and a b
ID: 1266015 • Letter: L
Question
left of the pulley N right of the pulley N A block of mass m_1 = 2.05 kg and a block of mass m_2 = 5.80 kg are connected by a massless string over a pulley in the shape of a solid disk having radius R = 0.250 m and mass M = 10.0 kg. These blocks are allowed to move on a fixed block-wedge of angle ? = 30.0. The coefficient of kinetic friction is 0.360 for both blocks. Draw free-body diagrams of both blocks and of the pulley. (a) Determine the acceleration of the two blocks. m/s^2 (b) Determine the tensions in the string on both sides of the pulley.Explanation / Answer
For the sum of forces of m1:
T1 - uk m1 g = m1a [1]
For the sum of forces of m2:
m2 g sin 30 - T2 - uk m2 g cos 30 = m2a [2]
For the sum of torques on the pulley,
(T2 - T1)R = 1/2 MR^2 (a/R)
(T2 - T1) = 1/2 M a [3]
As
m1 = 2.05 kg
m2 = 5.80 kg
R = 0.250 m
M = 10.0 kg
uk = 0.360
Plugging in the given and using these three equations,
a = 0.270 m/s^2 [ANSWER, PART A]
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T1 = 7.79 N [ANSWER, TENSION ON THE LEFT]
T2 = 9.13 N [ANSWER, TENSION ON THE RIGHT]
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