# PHYS101 The Laws of Motion Spring 2014

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1 The Laws of Motion 1. An object of mass m 1 = kg placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging object of mass m 2 = kg as shown in Figure 1. (a) Draw free-body diagrams of both objects. (b) Find the magnitude of the acceleration of the objects (c) Find the tension in the string. (d) If there is a friction µ k = 0.2 on the surface, repeat the part a,b, c, d (a) Free-Body Diagrams (Notice the chosen directions for acceleration) (b) Applying the second law for m 1 ΣF x = T = m 1 a (1) ΣF y = n m 1 g = 0 (2) for m 2 ΣF x = 0 (3) c 2014 Department of Physics, Eastern Mediterranean University Page 1 of 14

2 from (1) and (4) ΣF y = m 2 g T = m 2 a (4) m 2 g m 1 a = m 2 a m 1 = 55kg and m 2 = 59kg and g = 9.81ms 2 a = (c) To find the tension we may use (1) m 2 g = (m 1 + m 2 )a m a = 2 (m 1 + m 2 ) g (5) 59kg 59kg + 55kg 9.81ms 2 = 5.08ms 2 T = m 1 a = 55kg 5.08ms 2 = 279.4N (d) Free-Body diagram when friction exists between the surfaces for m 1 ΣF x = T f k = m 1 a ; f k = µ k n (6) from (6) and (7) ΣF y = m 1 g n = 0 n = m 1 g (7) T µ k m 1 g = m 1 a (8) for m 2 ΣF x = 0 (9) Now we add (8) and (10) ΣF y = m 2 g T = m 2 a (10) m 2 g µ k m 1 g = (m 1 + m 2 )a a = (m 2 µ k m 1 ) m 1 + m 2 g (11) a = and for tension we may use (10) (59kg kg) 9.81ms 2 55kg + 59kg = 4.13ms 2 T = m 2 (g a) = 59kg (9.81ms ms 2 ) = N c 2014 Department of Physics, Eastern Mediterranean University Page 2 of 14

3 (e) 2. Two objects are connected by a light string that passes over a frictionless pulley as shown in Figure P2. Assume the incline is frictionless and take m 1 = 2.00 kg, m 2 = 6.00 kg, and θ = 55 (a) Draw free-body diagrams of both objects. (b) Find the magnitude of the acceleration of the objects (c) Find the tension in the string. (d) Find the speed of each object 2.00 s after it is released from rest. (e) If there is a friction µ k = 0.2 on the surface, repeat the part a,b, c, d c 2014 Department of Physics, Eastern Mediterranean University Page 3 of 14

4 (a) Free-Body Diagram (Notice the chosen directions for acceleration) (b) Apply the second law for m 1 ΣF x = 0 (12) ΣF y = T m 1 g = m 1 a (13) for m 2 ΣF x = m 2 g sin θ T = m 2 a (14) Adding (13) and (14) we get a = ΣF y = n m 2 g cos θ = 0 (15) m 2 g sin θ m 1 g = (m 1 + m 2 )a a = (m 2 sin θ m 1 ) m 1 + m 2 g (16) (6kg sin 55 2kg) 2kg + 6kg (c) To calculate tension we use (13) 9.81ms 2 = 3.57ms 2 T = m 1 (g + a) = 2kg (9.81ms ms 2 ) = 26.8N (d) For both objects we have v i = 0 and we may use v f = v i + at (e) Free-Body Diagram: v f = at = 3.57ms s = 7.14ms 1 c 2014 Department of Physics, Eastern Mediterranean University Page 4 of 14

5 We now apply the second law to m 1 and m 2 when the incline is frictionless. for m 1 ΣF x = 0 (17) ΣF y = T m 1 g = m 1 a (18) and for m 2 ΣF x = m 2 g sin θ T f k = m 2 g ; f k = µ k n (19) ΣF y = n m 2 g cos θ = 0 n = m 2 g cos θ (20) from (19) and (20) m 2 g sin θ T µ k m 2 g cos θ = m 2 a (21) Adding (18) and (21) results in m 2 g sin θ µ k m 2 g cos θ m 1 g = (m 1 + m 2 )a a = from (18) a = m 2(sin θ µ k cos θ) m 1 m 1 + m 2 g (22) 6kg(sin cos 55) 2kg 2kg + 6kg 9.81ms 2 = 2.73ms 2 T = m 1 (g + a) = 2kg (9.81ms ms 2 ) = 25.08N It s speed at t = 2.00s is v f = at = 2.73ms s = 5.46ms 1 c 2014 Department of Physics, Eastern Mediterranean University Page 5 of 14

6 3. A woman at an airport is pulling her 20.0-kg suitcase at constant speed by pulling on a strap at an angle above the horizontal. She pulls on the strap with a 35.0-N force, and the friction force on the suitcase is 20.0N. (a) Draw a freebody diagram of the suitcase. (b) What angle does the strap make with the horizontal? (c) What is the magnitude of the normal force that the ground exerts on the suitcase? (a) Free-Body Diagram (b) Apply the second law to the suitcase but note that the right-hand side of the equation on the horizontal must be zero since the suitcase is moved at constant speed. ΣF x = F cos θ f k = 0 cos θ = f k F cos θ = 20.0N = 0.57 θ = arccos (0.57) = N (23) c 2014 Department of Physics, Eastern Mediterranean University Page 6 of 14

7 (c) Write the second law on the vertical ΣF y = n + F sin θ mg = 0 (24) n = (20.0kg)(9.81ms 2 ) 35.0N sin (55.15) = 167.5N 4. Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (Fig. P4). Suppose F = 68.0 N, m 1 = 12 kg, m 2 = 18 kg and the coefficient of kinetic friction between each block and the surface is (a) Draw a free-body diagram for each block. (b) Determine the acceleration of the system. (c) Determine the tension T in the rope. (a) Free-Body Diagram (b) Apply the second law to each block separately. For m 1 ΣF x = T f k1 = m 1a ; f k1 = µ kn 1 (25) from (25) and (26) we have ΣF y = n 1 m 1 g = 0 n 1 = m 1 g (26) T µ k m 1 g = m 1 a (27) for m 2 ΣF x = F T f k2 = m 2a ; f k2 = µ kn 2 (28) c 2014 Department of Physics, Eastern Mediterranean University Page 7 of 14

8 from (28) and (29) ΣF y = n 2 m 2 g = 0 n 2 = m 2 g (29) We now add (27) and (30) to obtain (c) From (27) F T µ k m 2 g = m 2 a (30) F µ k g(m 1 + m 2 ) = (m 1 + m 2 )a a = F µ kg(m 1 + m 2 ) m 1 + m 2 (31) a = 68.0N (0.100)(9.81ms 2 )(12.0kg kg) 12.0kg kg = 1.29ms 2 T = (µ k g + a)m 1 = [(0.100)(9.81ms 2 ) ms 2 ] 12.0kg = 27.3N 5. In Fig. 5, a tin of antioxidants (m 1 = 1.0kg) on a frictionless inclined surface is connected to a tin of corned beef (m 2 = 2.0kg). The pulley is massless and frictionless. An upward force of magnitude F = 6.0N acts on the corned beef tin, which has a downward acceleration of 5.5m/s 2. (a) What is the tension in the connecting cord? (b) What is the angle β? c 2014 Department of Physics, Eastern Mediterranean University Page 8 of 14

9 (a) To work out tension we apply the second law to m 2. ΣF x = 0 (32) ΣF y = m 2 g T F = m 2 a T = (g a)m 2 F (33) T = (9.81ms 2 5.5ms 2 )(2.0kg) 6.0N = 2.62N (b) To find the angle β apply the second law to m 1. ΣF x = T + m 1 g sin β = m 1 a sin β = m 1a T m 1 g (34) sin β = (1.0kg)(5.5ms 2 ) 2.62N 1.0kg 9.81ms 2 = 0.3 β = Arcsin(0.3) = 17.1 (35) Note that applying the second law for the vertical direction gives us information which is of no benefit for this particular problem. 6. In Figure P6, the pulleys and the cord are light, all surfaces are frictionless, and the cord does not stretch. c 2014 Department of Physics, Eastern Mediterranean University Page 9 of 14

10 (a) How does the acceleration of block 1 compare with the acceleration of block 2? (b) The mass of block 2 is 1.30 kg. Find its acceleration as it depends on the mass m 1 of block 1. Evaluate your answer for m 1 = 0.550kg. (c) What If? What does the result of part (b) predict if m 1 is very much less than 1.30 kg? (d) What does the result of part (b) predict if m 1 approaches infinity? (e) In this last case, what is the tension in the long cord? (f) Could you anticipate the answers to parts (c), (d), and (e) without first doing part (b)? Explain. (a) The moving pulley has an acceleration a 2. Since m 1 moves twice the distance this pulley moves in the same time, m 1 has twice the acceleration of the pulley. Hence a 1 = 2a 2 (b) from (36), (37), and (38) and the fact that a 1 = 2a 2 a 2 = m 2 g T 2 = m 2 a 2 (36) T 1 = m 1 a 1 (37) T 2 2T 1 = 0 (38) a 2 = m 2 m 2 + 4m 1 g (39) 1.30kg 1.30kg kg 9.81ms 2 = 3.64ms 2 (40) (c) If m 1 is very much less than m 2 = 1.30kg then the term in (39) including m 1 is negligible compared to m 2 and hence the acceleration of m 2 approaches g = 9.81ms 2. c 2014 Department of Physics, Eastern Mediterranean University Page 10 of 14

11 (d) If m 1 approaches infinity in (39) a 2 will approach zero. (e) If a 2 approaches zero in (36) T 2 = m 2 g and from (38) T 1 = 1 2 m 2g. (f) In case (c) since m 1 is negligible compared to m 2 we can simply conclude that m 2 freely falls with g. In (d) an infinitely massive block, m 1, does not let the other block move so it is predictable that the acceleration a 2 is zero. And finally in case (e) since the block 2 is stationary the forces,weight and tension, acting on it must balance and knowing the fact that T 2 = 2T 1 the same conclusion as we did in (e) comes out and that is T 1 = 1 2 m 2g. 7. What horizontal force must be applied to a large block of mass M shown in Figure 7 so that the tan blocks remain stationary relative to M? Assume all surfaces and the pulley are frictionless. Notice that the force exerted by the string accelerates m 2. F = (M + m 1 + m 2 )a (41) for m 1 T = m 2 a (42) c 2014 Department of Physics, Eastern Mediterranean University Page 11 of 14

12 for m 2 T m 1 g = 0 (43) Eliminating T we get m 2 a m 1 g = 0 a = m 1 m 2 g (44) for the entire system F = (M + m 1 + m 2 )( m 1 m 2 )g (45) 8. A car accelerates down a hill (see figure below), going from rest to 30.0m/s in 6.00 s. A toy inside the car hangs by a string from the car s ceiling. The ball in the figure represents the toy, of mass kg. The acceleration is such that the string remains perpendicular to the ceiling. (a) Determine the angle θ? (b) Determine the tension in the string? c 2014 Department of Physics, Eastern Mediterranean University Page 12 of 14

13 (a) Choosing x axis pointing down the slope v f = v i + at 30.0ms 1 = 0 + a(6.00ms 2 ) a = 5.00ms 2 ΣF x = mg sin θ = ma sin θ = a g (46) (b) sin θ = 5.00ms ms 2 = 0.51 θ = sin 1 (0.51) = 30.7 ΣF y = T mg cos θ = 0 (47) T = (0.100kg)(9.81ms 2 ) cos (30.7) = 0.843N 9. The two blocks (m = 16.0kg) and (M = 88.0kg) in Fig. 9 are not attached to each other.the coefficient of static friction between the blocks is µ k = 0.38, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force required to keep the smaller block from slipping down the larger block? for m F = (m + M)a a = F m + M (48) (49) ΣF x = F n = ma (50) ΣF y = m c 2014 Department of Physics, Eastern Mediterranean University Page 13 of 14

14 from (43), (44), and (45) F = (m + M)mg Mµ s (51) F = (16.0kg kg) 16.0kg 9.81ms kg 0.38 = 488N c 2014 Department of Physics, Eastern Mediterranean University Page 14 of 14

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