2. For every action force there is an equal and opposite reaction force!!

3. These Two forces are known as an action/reaction pair.These Two forces are known as an action/reaction pair. Gravity acts on an object and the object acts with equal force.Gravity acts on an object and the object acts with equal force.

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

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

6. A 60-kg athlete exerts a force on a 10-kg skateboard. If she receives an acceleration of 4 m/s 2, what is the acceleration of the skateboard? Force on runner = -(Force on board) m r a r = -m b a b (60 kg)(4 m/s 2 ) = -(10 kg) a b a = - 24 m/s 2 Force on Runner Force on Board

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

8. What is the tension T in the rope below if the block accelerates upward at 4 m/s 2 ? (Draw sketch and free-body.) 10 kg a = +4 m/s 2 Ta Tmg +  F x = m a x = 0 (No Motion)  F y = m a y = m a T - mg = m a mg = (10 kg)(9.8 m/s) = 98 N m a = (10 kg)(4 m/s) = 40 N - 98 N = 40 N T - 98 N = 40 N T = 138 N

9. 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 = m a to entire system (both masses). 12 N n (m 2 + m 4 )g  F x = (m 2 + m 4 ) a 12 N = (6 kg) a a =a =a =a = 12 N 6 kg a = 2 m/s 2

10. The two-body problem. 2 kg4 kg 12 N Now find tension T in connecting cord. Apply F = m a to the 2 kg mass where a = 2 m/s 2. T n m2 gm2 gm2 gm2 g  F x = m 2 a T = (2 kg)(2 m/s 2 ) T = 4 N

11. The two-body problem. 2 kg 4 kg 12 N Same answer for T results from focusing on 4-kg by itself. Apply F = m a to the 4 kg mass where a = 2 m/s 2.  F x = m 4 a 12 N - T = (4 kg)(2 m/s 2 ) T = 4 N 12 N n m2 gm2 gm2 gm2 g T

12. 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.  F x = (m 2 + m 4 ) a a = 6.53 m/s 2 n m2 gm2 gm2 gm2 g T m4 gm4 gm4 gm4 g T + a Note m 2 g is balanced by n. m 4 g = (m 2 + m 4 ) a (4 kg)(9.8 m/s 2 ) 2 kg + 4 kg a = = m 4 g m 2 + m 4 2 kg 4 kg

13. Now find the tension T given that the acceleration is a = 6.53 m/s 2. To find T, apply F = m a to just the 2 kg mass, ignoring 4 kg. T = (2 kg)(6.53 m/s 2 ) T = 13.1 N Same answer if using 4 kg. m 4 g - T = m 4 a T = m 4 (g - a ) = 13.1 N n m2 gm2 gm2 gm2 g T m4 gm4 gm4 gm4 g T + a 2 kg 4 kg

14. Vector Components So we have a person pulling a sled 30 o with respect to the horizontal at a force of 50 N.So we have a person pulling a sled 30 o with respect to the horizontal at a force of 50 N. We need to think of it like the sled being pulled vertically and horizontally at the same time, giving it both components.We need to think of it like the sled being pulled vertically and horizontally at the same time, giving it both components. Θ=30 o F=50N FyFy FxFx

15. Vector Components To solve for F x, we will use cosine because it is the adjacent side and we have the hypotenuse.To solve for F x, we will use cosine because it is the adjacent side and we have the hypotenuse. To solve for F y, use the same process but with sine.To solve for F y, use the same process but with sine. Θ=30 o F=50m/s FyFy FxFx

16. Angled Free-body Diagram 30 0 60 0 4 kgA A B B W = mg 30 0 60 0 BxBx ByBy AxAx AyAy 1. Draw and label sketch. 2. Draw and label vector force diagram. 3. Dot in rectangles and label x and y compo- nents opposite and adjacent to angles.

17. In the absence of friction, what is the acceleration down the 30 0 incline? 30 0 mg n n W mg sin 30 0 mg cos 30 0 +  F x = m a x mg sin 30 0 = m a a = g sin 30 0 a = (9.8 m/s 2 ) sin 30 0 a = 4.9 m/s 2

18. Summary 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.