HW #5-1-solutions

# HW #5-1-solutions - pasha(sep635 – HW 5-1 – Antoniewicz...

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Unformatted text preview: pasha (sep635) – HW 5-1 – Antoniewicz – (56445) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 7) 2.0 points A box of mass 40 kg hangs motionless from two ropes, as shown in the figure below. The angle is 36 ◦ . Choose the box as the system. The x axis runs to the right, the y axis runs up, and the z axis is out of the page. Rope 1 Rope 2 36 ◦ Is d ~ p/dt of the box zero or nonzero? 1. Zero correct 2. Unable to determine 3. Nonzero Explanation: The box is in static equilibrium, so the net force on the box is zero. 002 (part 2 of 7) 2.0 points Which of the free-body diagrams in the fol- lowing figure correctly describes the forces acting on the box? I ~ F T 2 (rope 2) ~ F grav (Earth) ~ F T 1 (rope 1) II ~ F T 2 (rope 2) ~ F grav (Earth) III ~ F T 2 (rope 2) ~ F T 1 (rope 1) IV ~ F T 2 (rope 2) ~ F grav (Earth) ~ F T 1 (rope 1) 1. I 2. II 3. III 4. IV correct Explanation: 003 (part 3 of 7) 2.0 points What is the y component of the gravitational force acting on the block? (A component can be positive or negative.) Use g = 9 . 8 m / s 2 . Correct answer:- 392 N. Explanation: Since gravity only acts along the y axis (in our coordinate system), we can write F grav ,y = F grav =- m g =- (40 kg)(9 . 8 m / s 2 ) =- 392 N . 004 (part 4 of 7) 3.0 points What is the y component of the force on the block due to rope 2? Correct answer: 392 N. Explanation: pasha (sep635) – HW 5-1 – Antoniewicz – (56445) 2 Since we’ve already found the force due to gravity acting downward on the box, this part is pretty simple. The box is not in motion, so the upward force must exactly balance the downward force. So the answer here is the same as for part 3, just with the opposite sign. 005 (part 5 of 7) 4.0 points What is the magnitude of ~ F T 2 ? Correct answer: 484 . 539 N. Explanation: Since we know the value of the y component of ~ F T 2 , finding the magnitude of the total tension in rope 2 is just trigonometry: ~ F T 2 cos 36 ◦ = 392 N ⇒ ~ F T 2 = 392 N cos 36 ◦ = 484 . 539 N . 006 (part 6 of 7) 4.0 points What is the x component of the force on the block due to rope 2? Correct answer: 284 . 805 N. Explanation: We use trigonometry again: F T 2 ,x = ~ F T 2 sin36 ◦ = (484 . 539 N) sin36 ◦ = 284 . 805 N . 007 (part 7 of 7) 3.0 points What is the x component of the force on the block due to rope 1?...
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## This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

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HW #5-1-solutions - pasha(sep635 – HW 5-1 – Antoniewicz...

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