Assume that the three blocks in the figure move on a frictionless surface. In the figure, m 1 = 1.10 kg, m 2 = 2.04 kg, and m 3 = 3.20 kg. The 42.1-N force acts as shown on the right block. Determine (a) the acceleration (m/s 2 ) of the blocks, (b) the tension (N) in the cord connecting the left and right blocks, and (c) the force (N) exerted on the center block by the left block. Hint: Draw a free-body diagram of each block
Given
masses m1 = 1.10 kg, m2 = 2.04 kg, m3 = 3.20 kg
force applied = 42.1 N
forces acting on m3 are applied force F (+ve) and tension in the chord (-ve)
forces acting on m1 are the tension (+ve) and the force applied by block 2 (-ve)
writing the equations for each block
m3
m1
m2
F-T = m3*a, T-f = m1*a
f = m2*a
solving equations
F - (m1*a+f) = m3*a
F - m1*a+m2*a = m3*a
F = a(m1+m3-m2)
a = F/(m1+m3-m2)
substitutign the values
a = 42.1/(1.10+3.20-2.04) m/s2
a = 18.62832 m/s2
the force exerted by block 2 on block 1 is f = m2*a = 2.04*18.62832 = 38 N in the netative x direction
tension T = m1*a+f
T = 1.10*18.62832+38 N
T = 58.491152 N
answers are
a) a = 18.62832 m/s2
b) T = 58.491152 N
c) f = 38 N
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