HW #11-solutions - wolz (cmw2833) HW #11 Antoniewicz...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: wolz (cmw2833) HW #11 Antoniewicz (57420) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 5.0 points A uniform horizontal beam of weight 400 N and length 4 . 49 m has two weights hanging from it. One weight of 300 N is located . 95637 m from the left end; the other weight of 200 N is located 0 . 95637 m from the right end. What must be the magnitude of the one additional force on the beam that will produce equilibrium? Correct answer: 900 N. Explanation: Let : W = 400 N , W 1 = 300 N , and W 2 = 200 N . For the beam to preserve translational equi- librium, the net force acting on the beam has to be zero, so the additional force should be F = W + W 1 + W 2 = 400 N + 300 N + 200 N = 900 N . 002 (part 2 of 2) 5.0 points At what distance from the left end must this force be applied for the beam to be in equilib- rium? Correct answer: 2 . 10182 m. Explanation: The beam also has to preserve its rotational equilibrium; i.e. , the net torque acting on the beam has to be zero, so F x = W L 2 + W 1 L 1 + W 2 L 2 x = W L 2 F + W 1 L 1 F + W 2 L 2 F = (400 N) (4 . 49 m) 2 (900 N) + (300 N) (0 . 95637 m) 900 N + (200 N) (3 . 53363 m) 900 N = 2 . 10182 m . 003 (part 1 of 4) 3.0 points Two weights are suspended as shown in the diagram. On the left, a weight of 79 N is suspended from its upper left by a string of tension T 1 angled at 44 with the horizontal. A string carrying tension T 2 is stretched be- tween them angled at 12 with the horizontal. On the right, the unknown weight W 2 is also suspended from its upper right by a string of tension T 3 angled at 50 with the horizontal. 79 N W 2 12 T 2 T 1 T 3 5 4 4 What is the tension T 1 ? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 145 . 822 N. Explanation: Let : W 1 = M 1 g = 79 N , 1 = 44 , 2 = 12 , and 3 = 50 . For the 79 N weight, horizontal equilibrium yields T 1 cos 1 = T 2 cos 2 T 2 = T 1 cos 1 cos 2 wolz (cmw2833) HW #11 Antoniewicz (57420) 2 and vertical equilibrium T 1 sin 1 = W 1 + T 2 sin 2 T 1 sin 1 = W 1 + T 1 cos 1 cos 2 sin 2 = W 1 + T 1 cos 1 tan 2 . T 1 sin 1- T 1 cos 1 tan 2 = W 1 T 1 = W 1 sin 1- cos 1 tan 2 = 79 N sin 44 - cos 44 tan12 = 145 . 822 N . 004 (part 2 of 4) 3.0 points What is the tension T 2 ? Correct answer: 107 . 239 N. Explanation: From the horizontal equation, T 2 = T 1 cos 1 cos 2 = (145 . 822 N) cos44 cos 12 = 107 . 239 N . 005 (part 3 of 4) 2.0 points What is the tension T 3 ? Correct answer: 163 . 188 N....
View Full Document

This note was uploaded on 05/02/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 8

HW #11-solutions - wolz (cmw2833) HW #11 Antoniewicz...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online