Question

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

42.1 N

Answer 1

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