This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY–302 K. Solutions for midterm test #2. Problem # 1 : First, let’s find how hard can a single laborer pull on a rope without slipping back. The diagram below shows the forces on the laborer: T 1 mg N L f L N R f R The laborer keeps walking forward at constant velocity, so all the forces on him must cancel, thus F net x = f L + f R − T 1 = 0 , F net y = N L + N R − mg. (1) As long as he does not slip, the friction forces acting on his feet are the static friction forces, which are limited by the respective normal forces according to f L ≤ μ dry s × N L and f R ≤ μ dry s × N R . (2) Consequently, the rope force T 1 is limited by T 1 = f L + f R ≤ μ dry s × ( N L + N R ) = μ dry s × mg. (3) If the rope pulls him with a force stronger than T max 1 = μ dry s × mg = 0 . 60 × 9 . 8 m / s 2 × 60 kg = 353 N , (4) he would slide back. And by the 3 rd Law of Newton, this is also the strongest force with which he can pull on the rope: If he pulls any harder, he would slip back. 1 Now consider the forces on the stone: Mg N f T To keep the stone sliding on the mud with a constant velocity, the net force on the stone must vanish, thus F net x = f − T = 0 , F net y = N − mg. (5) The friction force f is the sliding friction (also known as kinetic friction), so it’s governed by the normal force as f = μ mud k × N. (6) Therefore, the rope’s pull T must be T = f = μ mud k × N = mud × N = 0 . 1 × 7200 kg × 9 . 8 m / s 2 = 7060 N . (7) Comparing this required pulling force with a maximal force (4) provided by a single laborer, we immediately see that we need at least # min = T T max 1 = 7060 N 353 N = 20 laborers (8) to pull the stone. 2 To be precise, 20 is the minimal number of laborers needed to keep the stone moving once it has already been set in motion . To get the stone moving in the first place, we need a stronger force T start to overcome the static friction force on the stone. The static friction reaches a maximal value f max s = μ mud s × N = μ mud s × Mg = 0 . 15 × 7200 kg × 9 . 8 m / s 2 = 10600 N (9) before the stone begins to slide, so to get the stone moving we need to pull it with force T start ≥ T start min = f max s = 10600 N , (10) which takes at least # start min = T start min T max 1 = 30 laborers . (11) The bottom line is, it takes at least 30 laborers to get the stone moving, but once it moves, it takes only 20 laborers can keep it going at constant speed....
View
Full
Document
This note was uploaded on 11/11/2010 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Kaplunovsky
 Force

Click to edit the document details