the suspended 2.9 kg mass on the right is moving up, the 2.4 kg mass slides down
ID: 1391860 • Letter: T
Question
the suspended 2.9 kg mass on the right is moving up, the 2.4 kg mass slides down the ramp, and the suspended 8.7 kg mass on the left is moving down. coefficient of friction between the block and the ramp is 0.1. gravity is 9.80m/s^2
The pulleys are massless and frictionless.
a) what is the acceleration of the three block system in units of m/s^2
b) what is the tension of the cord connected to the 2.9 kg block? in units of N
c) what is the tension in the cord connected to the 8.7 kg block? in units of N
Explanation / Answer
force acting on the mass m1 at right side :
T1 - W1 = m1 a
T1 - m1 g = m1 a
where , m1 = 2.9 kg & g = 9.8 m/s2
inserting the values in above eq.
T1 - (2.9 kg) (9.8 m/s2) = (2.9 kg) a
T1 = 28.42 + 2.9 a { eq. 1 }
Force acting on the mass m2 at incline plane ::
W2x + T2 - us W2y - T1 = m2a
m2 g sin (theta) + T2 - us m2 g cos (theta) - T1 = m2 a
where, m2 = 2.4 kg
us = coefficient of friction = 0.1
angle = 25 degree
inserting all these values in above eq,
(2.4 kg) (9.8 m/s2) sin 25 + T2 - (0.1) (2.4 kg) (9.8 m/s2) cos 25 -T1 = (2.4 kg) a
9.93 + T2 - 2.13 - T1 = 2.4 a { eq. 2 }
now, inserting the value of T1 in eq.2,
9.93 + T2 - 2.13 - (28.42 + 2.9 a) = 2.4 a
- 20.62 + T2 = 2.4 a + 2.9 a
- 20.62 + T2 = 5.3 a { eq. 3 }
Force acting on the mass m3 at left side ::
W3 - T2 = m3 a
m3 g - T2 = m3 a
where, m3 = 8.7 kg
inserting the values in above eq,
(8.7 kg) (9.8 m/s2) - T2 = (8.7 kg) a
85.26 -T2 = 8.7 a { eq. 4 }
adding eq. 3 & 4,
85.26 -T2 = 8.7 a
- 20.62 + T2 = 5.3 a
solving above both eq,
64.64 = 14 a
a = 4.61 m/s2
using eq.1 & inserting the value of 'a' in it :
T1 = 28.42 + 2.9 a
T1 = 28.42 + 2.9 (4.61 m/s2)
T1 = 41.7 N
again, using eq.3
T2 = 5.3 a + 20.62
T2 = 5.3 (4.61 m/s2) + 20.62
T2 = 45 N
(a) the acceleration of the three block system is 4.61 m/s2
(b) the tension of the cord connected to the 2.9 kg block is 41.7 N
(c) the tension in the cord connected to the 8.7 kg block is 45 N
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