Question

Assume the three blocks (m. = 1.0 kg, m = 20 kg and m = 40 ko) portrayed in the figure below move on a frictionless surface and a force F: 36w acts as shown on the 4.0 kg block. (a Determine the acceleration given this system (in m/s2 to the right). m/s2 (to the right) b) Determine the tension in the cord connecting the 4.0 kg and the 1.0 kg blocks in N). Determine the force exerted by the 1.0 kg block on the 2.0 kg block (in N). N (a) What If? How would your answers to parts (a) and (b) of this problem change if the 2.0 kg block was now stacked on top of the 1.0 kg block? Assume that the 2.0 kg block sticks to and does not slide on the 1.0 kg block when the system is accelerated. (Enter the acceleration in m/s2 to the right and the tension in N.) acceleration m/s (to the right) tension

Answer 1

a)

acceleration of the system is given as

a = F/(m₁ + m₂ + m₃)

a = 36/(1 + 2 + 4)

a = 5.14 m/s²

b)

а T mga vр,

T = Tension force in the connecting cord

F = Applied force = 36 N

Consider the forces acting on m₃

Force equation for the motion of "m3 is given as

F - T = m₃ a

36 - T = 4 (5.14)

T = 15.44 N

c)

F₁₂ = Force by 1 kg block on 2 kg block

Consider the forces on 2 kg block,

Force equation for the motion of 2 kg block is given as

F₁₂ = m₂ a

F₁₂ = 2 (5.14)

F₁₂ = 10.3 N

d)

Since the three blocks move together

acceleration of the system still remains the same and is given as

a = F/(m₁ + m₂ + m₃)

a = 36/(1 + 2 + 4)

a = 5.14 m/s²

а T mga vр,

T = Tension force in the connecting cord

F = Applied force = 36 N

Consider the forces acting on m₃

Force equation for the motion of "m3 is given as

F - T = m₃ a

36 - T = 4 (5.14)

T = 15.44 N