= 1.49 m/s m. 2 kg. 2 kg

Transcription

1 5.6. Visualize: Please refer to Figure Ex5.6. Solve: For the diagra on the left, three of the vectors lie along the axes of the tilted coordinate sste. Notice that the angle between the 3 N force and the -axis is the sae 0 b which the coordinates are tilted. Appling Newton s second law, ( F ) net x 5 N 1 N ( 3sin0 ) N ax = = = 1.49 /s kg a F net = ( ) = ( ).8 N 3cos0 N = 0 /s kg For the diagra on the right, the -newton force in the first quadrant akes an angle of 15 with the positive x-axis. The other -newton force akes an angle of 15 with the negative -axis. The accelerations are a a F ( ) net x x = = = 0.8 /s F ( cos15 ) N + ( sin15 ) N 3 N kg N + ( sin15 ) N ( cos15 ) N = 0 /s kg net = ( ) =

2 5.14. Model: We assue that the passenger is a particle acted on b onl two vertical forces: the downward pull of gravit and the upward force of the elevator floor. Visualize: Please refer to Figure Ex5.14. The graph has three segents corresponding to different conditions: (1) increasing velocit, eaning an upward acceleration, () a period of constant upward velocit, and (3) decreasing velocit, indicating a period of deceleration (negative acceleration). Solve: Given the assuptions of our odel, we can calculate the acceleration for each segent of the graph and then appl Equation The acceleration for the first segent is v v a = /s 0 /s = = 4 /s t t s 0 s 1 0 w = w a 4 /s + = ( ) + = ( )( g ) + 4 g = app kg 9.8 /s N 9.8 /s 98. For the second segent, a = 0 /s and the apparent weight is For the third segent, w app 0 /s = w 1 + g = g = ( 75 kg)( 9.8 /s )= 740 N v3 v 0 /s 8 /s a = = = /s t3 t 10 s 6 s w = w + /s app 1 = ( 75 kg)( 9.8 /s )( 1 0. )= 590 N 9.8 /s Assess: As expected, the apparent weight is greater than noral when the elevator is accelerating upward and lower than noral when the acceleration is downward. When there is no acceleration the weight is noral. In all three cases the agnitudes are reasonable, given the ass of the passenger and the accelerations of the elevator.

3 5.16. Model: We assue that the truck is a particle in equilibriu, and use the odel of static friction. Visualize: Solve: The truck is not accelerating, so it is in equilibriu, and we can appl Newton s first law. The noral force has no coponent in the x-direction, so we can ignore it here. For the other two forces: ( ) = = = = = = ( )( ) F ΣF f w 0 N f w gsinθ 4000 kg 9.8 /s ( sin15 )= 10,145 N net x x s x s x Assess: The truck s weight (g) is roughl 40,000 N. A friction force that is 5% of the truck s weight sees reasonable.

4 5.33. Model: We can assue the foot is a single particle in equilibriu under the cobined effects of gravit, the tensions in the upper and lower sections of the traction rope, and the opposing traction force of the leg itself. We can also treat the hanging ass as a particle in equilibriu. Since the pulles are frictionless, the tension is the sae everwhere in the rope. Because all pulles are in equilibriu, their net force is zero. So the do not contribute to T. Visualize: Solve: (a) Fro the free-bod diagra for the ass, the tension in the rope is T w g = = = ( )( )= 6 kg 9.8 /s 58.8 N (b) Using Newton s first law for the vertical direction on the pulle attached to the foot, ( Fnet ) = ΣF = Tsinθ Tsin15 wfoot = 0 N Tsin15 + w g ( )( 4 kg 9.8 /s ) foot foot sinθ = = sin15 + = = = T T 58.8 N θ = sin = (c) Using Newton s first law for the horizontal direction, ( Fnet ) = ΣFx = Tcosθ + Tcos15 Ftraction = 0 N x traction = cosθ + cos15 = cos cos15 F T T T( ) = (58.8 N)( ) = (58.8 N)(1.344) = 79.0 N Assess: Since the tension in the upper segent of the rope ust support the foot and counteract the downward pull of the lower segent of the rope, it akes sense that its angle is larger (a ore direct upward pull). The agnitude of the traction force, roughl one-tenth the weight of a huan bod, sees reasonable.

5 5.51. Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we will also use the odel of kinetic friction. Visualize: Solve: The noral force due to the wall, which is perpendicular to the wall, is here to the right. The box slides down the wall at constant speed, so r r a = 0 and the box is in dnaic equilibriu. Thus, F r net = 0 r. Newton s second law for this equilibriu situation is (F net ) x = 0 N = n F push cos45 (F net ) = 0 N = f k + F push sin45 w = f k + F push sin45 g The friction force is f k = µ k n. Using the x-equation to get an expression for n, we see that f k = µ k F push cos45. Substituting this into the -equation and using Table 5.1 to find µ k = 0.0 gives, µ k F push cos45 + F push sin45 g = 0 N F = push ( )( ) + = kg 9.8 /s g + = 3.1 N µ cos45 sin cos45 sin 45 k

6