Example 6 Practice Problems

1. Given:   F_app = 30 N [forward],     m_A = 25 kg,     m_B = 15 kg,     a =0.75 m/s² [forward]
   Required:   The forces that the boxes exert on each other.
   Analysis:   Calculate the net force on box B using F_net = m_Ba and use Newton's Third Law to infer the force exerted by box B on box A
   Solution:   Net force on box B is F_net = m_Ba = 15x0.75 = 11.0 N [forward]
   Statement:   The net force on box B is 11N [forward]. This is the force exerted by box A, and by Newton's Third Law this is the force exerted by box B on box A 11 N [backward].

2. Data: The mass of each bucket is 5.0 kg. The action-reaction force between the buckets is 60 N N.
   Required:   acceleration (a) and the applied force (F_app)
   Analysis:
   • set [up] as the positve direction. Use the action-reaction force and the force of gravity to find the net force on the bottom bucket.
   • Use the net force and the mass of the bottom bucket to find the acceleration of the bottom bucket
   • The acceleration of the top bucket is the same as the bottom bucket
   • Use this accleration to calculate the net force on the top bucket
   • Calculate the applied force knowing the net force and the action-reaction force of 60 N
   Solution
   Force on the bottom bucket is F_g is 5.0 x 9.8 = 49 N [downward]
   Since the action-reaction force is 60N, the net force on the bottom bucket is 11N [up] Subtract the forces because they are in opposite directions.
   Using F_net = ma, the acceleration becomes 11/5 or 2.2 m/s² [up].
   Thus for the top bucket F_net = ma, which becomes 5 x 2.2 = 11N [up]. Note; be careful with these numbers, the problem has the simplicication that both buckets weigh the same, this makes answers to become numerically the same.
   The applied force must be larger than the action-reaction force by 11N. Thus the applied force is 60 N + 11 N = 71 N [up]
   Statement:   The acceleration is 2.2 m/s² [up] and the applied force is 71 N [up]