Newton's 3rd Law!! For every action force there is an equal and opposite reaction force!! For every action force there is an equal and opposite reaction force!!

For every action force there is an equal and opposite reaction force!! Newton's 3rd Law!! For every action force there is an equal and opposite reaction force!! These Two forces are known as an action/reaction pair. Gravity acts on an object and the object acts with equal force.

Newton’s Third Law Action-Reaction Pairs The hammer exerts a force on the nail to the right. The nail exerts an equal but opposite force on the hammer to the left.

Acting and Reacting Forces Use the words by and on to study action/reaction forces below as they relate to the hand and the bar: Action The action force is exerted by the _____ on the _____. hands bar Reaction The reaction force is exerted by the _____ on the _____. bar hands

Force on runner = -(Force on board) A 60-kg athlete exerts a force on a 10-kg skateboard. If she receives an acceleration of 4 m/s2, what is the acceleration of the skateboard? Force on runner = -(Force on board) mr ar = -mb ab (60 kg)(4 m/s2) = -(10 kg) ab Force on Board Force on Runner a = - 24 m/s2

Applying Newton’s Law Read, draw, and label problem. Draw free-body diagram for each body. Choose x or y-axis along motion and choose direction of motion as positive. Write Newton’s law for both axes: SFx = m ax SFy = m ay Solve for unknown quantities.

What is the tension T in the rope below if the block accelerates upward at 4 m/s2? (Draw sketch and free-body.) 10 kg SFx = m ax = 0 (No Motion) T a SFy = m ay = m a T - mg = m a a = +4 m/s2 mg = (10 kg)(9.8 m/s) = 98 N T mg + m a= (10 kg)(4 m/s) = 40 N T = 138 N T - 98 N = 40 N

First apply F = ma to entire system (both masses). Two-Body Problem: Find tension in the connecting rope if there is no friction on the surfaces. 2 kg 4 kg 12 N Find acceleration of system and tension in connecting cord. First apply F = ma to entire system (both masses). n SFx = (m2 + m4) a 12 N 12 N = (6 kg) a (m2 + m4)g a = 12 N 6 kg a = 2 m/s2

Apply F = m a to the 2 kg mass where a = 2 m/s2. The two-body problem. 2 kg 4 kg 12 N Now find tension T in connecting cord. Apply F = m a to the 2 kg mass where a = 2 m/s2. n SFx = m2 a T m2 g T = (2 kg)(2 m/s2) T = 4 N

Apply F = m a to the 4 kg mass where a = 2 m/s2. The two-body problem. 12 N Same answer for T results from focusing on 4-kg by itself. 2 kg 4 kg Apply F = m a to the 4 kg mass where a = 2 m/s2. 12 N n m2 g T SFx = m4 a 12 N - T = (4 kg)(2 m/s2) T = 4 N

First apply F = m a to entire system along the line of motion. Find acceleration of system and tension in cord for the arrangement shown. First apply F = m a to entire system along the line of motion. 2 kg 4 kg SFx = (m2 + m4) a Note m2g is balanced by n. n m2 g T m4 g + a m4g = (m2 + m4) a (4 kg)(9.8 m/s2) 2 kg + 4 kg a = = m4g m2 + m4 a = 6.53 m/s2

Now find the tension T given that the acceleration is a = 6.53 m/s2. To find T, apply F = m a to just the 2 kg mass, ignoring 4 kg. 2 kg 4 kg n m2 g T m4 g + a T = (2 kg)(6.53 m/s2) T = 13.1 N Same answer if using 4 kg. m4g - T = m4 a T = m4(g - a) = 13.1 N

Summary Newton’s First Law: An object at rest or an object in motion at constant speed will remain at rest or at constant speed in the absence of a resultant force. Newton’s Second Law: A resultant force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass. Newton’s Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs.