Elasticity - sanchez (ds28677) statics and elasticity...

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

View Full Document Right Arrow Icon
sanchez (ds28677) – statics and elasticity problems – Turner – (58220) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A cylindrical stone column oF diameter 2 R = 1 . 31 m and height H = 2 . 08 m is transported in standing position by a dolly. a side view When the dolly accelerates or decelerates slowly enough, the column stands upright, but when the dolly’s acceleration magnitude exceed a critical value a c , the column top- ples over. (±or a > + a c the column topples backward; For a < - a c the column toppes Forward.) Calculate the magnitude oF the critical ac- celeration a c oF the dolly. The acceleration oF gravity is 9 . 8 m / s 2 . Correct answer: 6 . 17212 m / s 2 . Explanation: Let : g = 9 . 8 m / s 2 , R = 0 . 655 m , and H = 2 . 08 m . In the non-inertial Frame oF the accelerating dolly, the column is subject to the horizontal inertial Force v F in = - mvg . Together, the gravity and the inertial Force combine into the apparent weight Force v W app = m ( vg - va ) in the direction θ = arctan p a g P From the vertical. ±rom the torque point oF view, this appar- ent weight Force applies at the center oF mass oF the column. The column is stable in the vertical position when the line oF this Force goes through the column’s base CM W app but when this line misses the base, the column topples over CM W app ±or the critical acceleration a c , the line goes through the edge oF the base, hence the direc- tion oF the apparent weight Force must deviate From the vertical by the angle θ c where tan θ c = R h cm = 2 R H . Consequently, the critical acceleration oF the dolly is a c = g tan θ c = g 2 R H = (9 . 8 m / s 2 ) 2 (0 . 655 m) 2 . 08 m = 6 . 17212 m / s 2 . 002 10.0 points The system shown in the fgure is in equilib- rium. A 15 kg mass is on the table. A string attached to the knot and the ceiling makes an angle oF 68 with the horizontal. The coe²- cient oF the static Friction between the 15 kg mass and the surFace on which it rests is 0 . 42.
Background image of page 1

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

View Full DocumentRight Arrow Icon
sanchez (ds28677) – statics and elasticity problems – Turner – (58220) 2 15 kg m 68 What is the largest mass m can have and still preserve the equilibrium? The accelera- tion of gravity is 9 . 8 m / s 2 . Correct answer: 15 . 5931 kg. Explanation: Let : M = 15 kg , m = 15 . 5931 kg , and θ = 68 . For the system to remain in equilibrium, the net forces on both M and m should be zero, so the tension in the rope has an upper bound value T max , where T max cos θ = μM g (1) T max = μM g cos θ = (0 . 42) (15 kg) (9 . 8 m / s 2 ) cos 68 = 164 . 813 N . For
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/07/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 7

Elasticity - sanchez (ds28677) statics and elasticity...

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