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**Unformatted text preview: **own as the +y direction) and obtain 1 PHY2048 Spring 2014 HW3 mC g – TBC = mC a. This leads to TBC = 36.8 N. (b) We use Eq. 2- 15 (choosing rightward as the +x direction): Δx = 0 + 1 at2 = 0.191 2
m. 5P52. The free- body diagram for the man is: And for the sandbag: Viewing the man- rope- sandbag as a system means that we should be careful to choose a consistent positive direction of motion (though there are other ways to proceed, say, starting with individual application of Newton’s law to each mass). We take down as positive for the man’s motion and up as positive for the sandbag’s motion and, without ambiguity, denote their acceleration as a. The net force on the system is the difference between the weight of the man and that of the sandbag. The system mass is msys = 85 kg + 65 kg = 150 kg. Thus, Eq. 5- 1 leads to 2 PHY2048 Spring 2014 HW3 (85 kg)(9.8 m/s2 ) − (65 kg)(9.8 m/s 2 ) = msys a which yields a = 1.3 m/s2. Since the system starts from rest, Eq. 2- 16 determines the speed (after traveling Δ y = 10 m) as follows: v = 2aΔy = 2(1.3 m/s2 )(10 m) = 5.1 m/s. 5P53. We apply Newton’s second law first to the three blocks as a single system and then to the individual blocks. The +x direction is to the right in Fig. 5- 48. (a) With msys = m1 + m2 + m3 = 67.0 kg, we apply Eq. 5- 2 to the x motion of the !
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system, in which case, there is only one force T3 = + T3 " . Therefore, i T3 = msys a ⇒ 65.0 N = (67.0 kg)a which yields a = 0.970 m/s2...

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