MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 2/7/06 PHYS103: Discusion 7 19. Two blocks are fastened to the ceiling of an elevator as in Figure P4.19. The elevator accelerates upward at 2.00 m/s 2 . Find the tension in each rope. (Draw FBD for each mass first) Figure P4.19 4.19 Fre-body diagrams of the two blocks are shown at the right. Note that each block experiences a downward gravitational forceF g =98.0 N. Also, each has the same upward aceleration as the elevator, y =+2.00 2 Aplying Newton’s 2 nd law to the lower block : ΣF y ⇒T ler −F g =a y or T lower = g +ma98.0 N10.0 k () +2.00 ms 2 () =118 N Then, aplying the 2 nd law to the uper block ΣF y upper −T ler F g =a y or upper = lower + gy 118 N+98.010.0 kg () +2.00 s 2 () =236 N MT117MT114 MT117MT114 MT117MT114 MT113 MT117MT114 MT117MT114 MT117MT114 MT117MT114 26. Two packing crates of masses 10.0 kg and 5.00 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.26. The 5.00-kg crate lies on a smooth incline of angle 40.0°. Find the acceleration of the 5.00-kg crate and the tension in the string. (Again draw 2 FBD) Figure P4.26 4.26 Let m 1 =10.0 kg, 2 =5.00 kg, and θ=40.0°. Aplying the second law to each object gives 1 a−T (1) and 2 = 2 siθ (2) Ading these equations yields m 1 − 2 + g , or a= 10.0 k5.00 k () sin40.0° 15.0 9.80 ms 2 () =4.43 s 2 Then, Equation (1) yields =mg−10.0 kg ) 9.80−4.43 () s 2 =53.7 N MT113 ° MT117MT114 MT117MT114 MT117MT114 MT117MT114 41. The coefficient of static friction between the 3.00-kg crate and the 35.0° incline of Figure P4.41 is 0.300. What minimum force F must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline? Figure P4.41 4.41 The normal force acting on the crate is given by =F+gsθ. The net force tending to move the crate down the incline is − s , where f s is the force of static friction betwen the crate and the incline. If the crate is in equilibrium, then i s =0, so that s =F g θ But, we also know f s ≤µ s F+ () Therefore, we may write ginθ s mcsθ, or F≥ iθ s −c = 35.0° 0.300 −° 3.00 kg () 9.80 ms 2 () =32.1 N 64. A 5.0-kg penguin sits on a 10-kg sled, as shown in Figure P4.64. A horizontal force of 45 N is applied to the sled, but the penguin attempts to impede the motion by holding onto a cord attached to a wall. The coefficient of kinetic friction between the sled and the snow, as well as that between the sled and the penguin, is 0.20. (a) Draw a free-body diagram for the penguin and one for the sled, and identify the reaction force for each force you include. Determine (b) the tension in the cord and (c) the acceleration of the sled. Figure P4.64 4.64 (a) Force diagrams for penguin and sled are shown. The primed forces are reaction forces for the coresponding unprimed forces. MT162 MT162 MT162 MT162 MT162 MT162 MT162MT162 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 MT117MT114 (b) The weight of the penguin is 49 N, and hence the normal force exerted on him by the sled, n 1 , is also 49 N. Thus, the friction force acting on the penguin is: 1 =µ k 0.2049 N () =9.8 Since the penguin is in equilibrium, the tension in the cord atached to the wal and the friction force must be equal: =9.8 N (c) The normal force exerted on the sled by the Earth is the weight of the penguin (49 N) plus the weight of the sled (98 N). Thus, the net normal force, 2 equals 147 N, and the friction force betwen sled and ground is: f 2 µ k =0.20147 N () =29.4 N Aplying the second law to the horizontal motion of the sled gives: 45 N−f 1 ′ 2 =10 k () or 0.58 2 Sridhara Dasu Dis07Solution