As depicted in the figure, a uniform rod of length 30 cm and mass 3.0 kg is pulled by strings subjected to constant forces of 20 N and 32 N . Determine the force exerted by the 20 cm part of the rod on the 10 cm part. All surfaces are smooth, and the strings and pulleys are light.

Step-by-step solution

Total mass of the rod, M = 3.0 kg Total length of the rod, L = 30 cm The net force acting on the rod is: F_ net = 32 - 20 = 12 N towards the right. The acceleration of the rod is: a = F_ net M = 12 3.0 = 4 m/s ^2 Since the rod is uniform, the mass of the 10 cm part is: m₁ = 10 30 3.0 = 1.0 kg Let F be the force exerted by the 20 cm part on the 10 cm part. The forces acting on the 10 cm part in the horizontal direction are 20 N to the left and F to the right. Applying Newton's second law to the 10 cm part: F - 20 =