Homework 6

# Homework 6 - Gozick, Brandon Homework 6 Due: Sep 29 2006,...

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Gozick, Brandon – Homework 6 – Due: Sep 29 2006, 5:00 pm – Inst: D Weathers 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Joe pushes down the length oF the handle oF a(n) 12 . 2 kg lawn spreader. The handle makes an angle oF 45 . 9 with the horizontal. Joe wishes to accelerate the spreader From rest to 1 . 33 m / s in 1 . 3 s. What Force must Joe apply to the handle? Correct answer: 17 . 9355 N. Explanation: The horizontal component oF the Force is F h = F cos θ. Let v be the fnal velocity oF the spreader. According to Newton’s second law, F h = ma h so F cos θ = m Δ v h Δ t F = m Δ v h Δ t cos θ = mv t cos θ = (12 . 2 kg)(1 . 33 m / s) (1 . 3 s) cos45 . 9 = 17 . 9355 N keywords: 002 (part 1 oF 1) 10 points A 56 kg boy and a 35 kg girl use an elastic rope while engaged in a tug-oF-war on a Frictionless icy surFace. IF the acceleration oF the girl toward the boy is 1 . 5 m / s 2 , determine the magnitude oF the acceleration oF the boy toward the girl. Correct answer: 0 . 9375 m / s 2 . Explanation: According to Newton’s Third Law oF Mo- tion the Force the boy exerts on the girl is the same as the Force the girl exerts on the boy, so F g = F b m g a g = m b a b a b = m g a g m b keywords: 003 (part 1 oF 2) 10 points An object in equilibrium has three Forces ex- erted on it. A(n) 29 N Force acts at 73 . 9 , and a(n) 48 N Force acts at 24 . 7 . What is the direction oF the third Force? (Consider all angles to be measured counter- clockwise From the positive x -axis.) Correct answer: 222 . 855 . Explanation: Basic Concepts: Choose a coordinate system with the pos- itive x -axis representing 0 and the positive y-axis representing 90 . Any Force F has an x -component oF F x = F cos θ and a y -component oF F y = F sin θ where θ is measured From the x -axis. Solution: Since F 1 and F 2 act up and to the right, the Force F must act down and to the leFt, so F lies in the third quadrant. θ θ θ 1 1 2 2 ref F F F ±or vertical equilibrium, F yNet = Σ F up - Σ F down = 0 F 1 y + F 2 y - F y = 0 F 1 sin θ 1 + F 2 sin θ 2 - F sin θ = 0 F sin θ ref = F 1 sin θ 1 + F 2 sin θ 2 (1) ±or horizontal equilibrium, F xNet = Σ F right - Σ F left = 0 F 1 x + F 2 x - F x = 0

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Gozick, Brandon – Homework 6 – Due: Sep 29 2006, 5:00 pm – Inst: D Weathers 2 F 1 cos θ 1 + F 2 cos θ 2 - F cos θ = 0 F cos θ ref = F 1 cos θ 1 + F 2 cos θ 2 (2) Dividing (1) by (2), we have F sin θ F cos θ = F 1 sin θ 1 + F 2 sin θ 2 F 1 cos θ 1 + F 2 cos θ 2 tan θ = F 1 sin θ 1 + F 2 sin θ 2 F 1 cos θ 1 + F 2 cos θ 2 θ ref = arctan µ F 1 cos θ 1 + F 2 θ 2 F 1 cos θ 1 + F 2 cos θ 2 Since the reference angle is in the third quadrant, and F 1 sin θ 1 + F 2 sin θ 2 = (29 N)sin73 . 9 + (48 N)sin24 . 7 = 47 . 9202 N and F 2 cos θ 1 + F 2 cos θ 2 = (29 N)cos73 . 9
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## This note was uploaded on 09/12/2008 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.

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Homework 6 - Gozick, Brandon Homework 6 Due: Sep 29 2006,...

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