Discussion Session 6 Newton s Second & Third Laws Week 07. The Plan

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1 PHYS 100 Discussion Session 6 Newton s Second & Third Laws Week 07 The Plan This week we use Newton s Third Law ( F A on B = F B on A ) to relate the forces between two different objects. We can use this principle, along with Newton s Second Law ( F net on A = m A a A ), to find the acceleration of an object in contact with other objects. Alternatively, we can find the forces between the objects if their acceleration is known. Your group will begin work today with a warm-up exercise having two objects that are not touching. Instead, they re connected with a string. This will give your group practice in dealing with multiple objects before we add the complication of using Newton s Third Law. It s also a nice review of applying Newton s Second Law from last week. After the warm-up, your group will tackle several problems with blocks touching one another while accelerating. You ll use combinations of Newton s Second and Third Laws to find the contact forces between objects and/or the objects acceleration. Newton s Third Law appears deceptively simple when stated mathematically. Here s an alternate plain English statement that might prove helpful to you in learning to use it: Newton s Third Law F A on B = F B on A When A pushes on B, B pushes back (on A) with equal strength. back ï opposite direction (that s what the minus sign tells you) equal strength ï same magnitude (that s what the equal sign tells you) 1

2 DQ1) This problem is a warm-up using Newton s Second Law, only now you have two connected objects instead of one isolated object. m m1 Blocks of mass m1 and m are suspended by a rope hung over two ideal pulleys, as shown in the figure. Initially, you are holding them so that the system does not move. At t = 0, you let go and the blocks accelerate. a) Work with your group to draw a Free Body Diagram for each object, indicating on each diagram the direction of positive motion. (make sure positive motion for one block corresponds to positive motion for the other block!) Check your diagrams with the rest of your group before proceeding. b) Working with the rest of your group, derive expressions for T, the tension in the string, and a, the acceleration of the two blocks, in terms of m 1, m, and g. (Question: how did I know that both blocks would have the same acceleration?) Substitute the expression for the acceleration into the m 1 equation to determine T. c) Check some limits to see if your expressions make sense: (i) In the limit that m 1 = m, what happens to a? Does this make sense? In this limit, a goes to zero and T goes to mg! That makes sense; if the masses are equal, one can t pull the other down. (ii) In the limit that m 1 is much larger than m, what happens to a? Does this make sense? In this limit, a goes to g and T goes to 0! That makes sense; m approaches free fall as m 1 goes to 0.

3 DQ) Three blocks (masses 3kg, 7kg, and 5kg) lay on a flat frictionless floor. A force with magnitude 5 N is applied to the 3 kg mass as shown in the figure. The three blocks accelerate to the right. 5N 3kg 7kg 5kg Your group s task is to find the force from the 5kg block on the 7 kg block. The following steps can help guide your group through this thinking process. a) Can you choose a system such that you can determine the acceleration of the 7 kg block using just one free-body diagram? If so, what is it? You bet you can!! The acceleration of the 3 kg block is the same as the acceleration of all the blocks together. Therefore, just choose the system to be the three blocks considered as one! b) What system would you choose to calculate the force The 5 kg block exerts on the 7 kg block? Draw the free-body diagram. You can use either the 5kg block alone OR the combination of the 7kg block and the 3 kg block. Why? Newton s Third Law says that the magnitudes of F 5on7 and F 7on5 are the same! With these pictures, you should be in a position to determine the force the 5 kg block exerts on the 7 kg block. On the next page, we show the solution map for this problem. You may find it helpful in solving this problem. 3

4 Calculate F5on7! Write down Newton s Second Law for the 3-block system: F = (m 1 + m + m 3 ) a Solve for a: a = F / (m 1 + m + m 3 ) = 5N / 15 kg a = 5/3 m/s Write down Newton s Second Law for the -block system: F F 5on7 = (m 1 + m ) a Solve for F 5on7 : F 5on7 = F (m 1 + m ) a = 5N 10kg*(5/3)m/s F 5on7 = 5/3 N 4

5 DQ3) Two crates are at rest in an elevator that moves upward with constant acceleration a as shown in the figure below. Work with your fellow group members to answer the following questions about this situation. a Elevator Assume m 1, m, and a are known. m m 1 g a. Find the force from m 1 on m (F 1on ) in terms of m 1, m, g, and a, as needed. Write down Newton s Second Law for the FBD shown: F 1on m g = m a Solve for F 1on F 1on = m (a + g) Direction = UP b. Find the force from m on m 1 (F on1 ) in terms of m 1, m, g, and a, as needed. Easy Way: Newton s Third Law F on1 = F 1on F on1 = m (a + g) Direction = DOWN c. Find the force from the floor on m 1 (N Floor on 1 ) in terms of m 1, m, g, and a, as needed. Write down Newton s Second Law for the FBD shown: N Flooron1 F 1on m 1 g = m 1 a Solve for N Flooron1 N Flooron1 = F 1on + m 1 (g + a) N Flooron1 = = m (a + g) + m 1 (g + a) N Flooron1 = (m 1 +m )(g + a) NOTE: This is the same result you would have gotten if you had chosen m 1 +m as your system!! 5

6 DQ4) Here we have a two-dimensional problem that can be done using the same procedure for the preceding one-dimensional questions. Two blocks (masses m 1 and m ) are on a ramp inclined at an angle θ from the horizontal. A force with magnitude F is applied to the block with mass m 1 up the ramp. The two blocks accelerate up the ramp. Talk with your fellow group members to determine a strategy for finding the force from m on m 1 (F on1 ) in terms of m 1, m, θ and F. Here is a solution map for the problem you may feel useful in designing and implementing your strategy. Determine a using both blocks as the system Write down Newton s Second Law for the two-block system: F (m 1 + m ) g sinθ = m a a = (F/(m 1 + m )) g sinθ Write down Newton s Second Law for the single block (m ) system: F 1on m g sinθ = m a F 1on = m g sinθ + m *((F/(m 1 + m )) g sinθ) F 1on = (m /(m 1 + m )) F Newton s Third Law implies the magnitude of F on1 is the same as the magnitude of F 1on 6

7 Formula Sheet Definitions Position Velocity Acceleration x v = dx dt a = = dv dt d x dt Constant Acceleration v v at = 0 + x = x + v t + at ( ) v = v + a x x 0 0 Relative Motion va, B = va, E + ve, B v = v E, B B, E Newton s Laws F = F = ma net i i Constants and Conversions m g = = 3 ft s s 1 mile = km Quadratic Formula If ± 4 ax + bx + c = 0 then x = b b ac a 7