# hjm - Version 108/ABCDA final Demkov (59910) This print-out...

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

Version 108/ABCDA – fnal – Demkov – (59910) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A uniForm bar oF length L and weight W is attached to a wall with a hinge that exerts on the bar a horizontal Force H x and a vertical Force H y . The bar is held by a cord that makes a 90 angle with respect to bar and angle θ with respect to wall. The acceleration oF gravity g = 9 . 8 m / s 2 . L 90 W θ What is the magnitude oF the horizontal Force H x on the pivot? 1. H x = 1 2 W tan θ 2. H x = 1 2 W cos θ 3. H x = 1 2 W sin θ cos θ correct 4. H x = 1 2 W sin θ 5. H x = 1 2 W cos 2 θ 6. H x = 1 2 W sin 2 θ Explanation: Analyzing the torques on the bar, with the hinge at the axis oF rotation, we have s τ = LT p L 2 sin θ P W = 0 , so, T = 1 2 W sin θ . Analyzing the Force on the bar, we have s F x = H x T cos θ = 0 . Put T into this equation and get H x = p 1 2 W sin θ P cos θ . 002 10.0 points A triangular wedge 7 m high, 14 m base length, and with a 13 kg mass is placed on a Frictionless table. A small block with a 8 kg mass (and negligible size) is placed on top oF the wedge as shown in the fgure below. 13 kg 14 m 7 m 8 kg Δ X wedge 13 kg 14 m All surFaces are Frictionless, so the block slides down the wedge while the wedge slides sidewise on the table. By the time the block slides all the way down to the bottom oF the wedge, how Far Δ X wedge does the wedge slide to the right?

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

View Full Document
Version 108/ABCDA – fnal – Demkov – (59910) 2 1. 4.72727 2. 4.66667 3. 5.71429 4. 4.95238 5. 4.58824 6. 3.94737 7. 3.61111 8. 5.89474 9. 5.33333 10. 4.11765 Correct answer: 5 . 33333 m. Explanation: Let : M = 13 kg , m = 8 kg , L = 14 m , and H = 7 m . Consider the wedge and the block as a two- body system. The external forces acting on this system — the weight oF the wedge, the weight oF the block and the normal Force From the table — are all vertical, hence the net hor- izontal momentum oF the system is conserved, P wedge x + P block x = constant . ±urthermore, we start From rest = center- oF-mass is not moving, and thereFore the X coordinate oF the center-oF-mass will remain constant while the wedge slides to the right and the block slides down and to the leFt, X cm = mX block + M X wedge m + M = constant . Note: Only the X coordinate oF the center- oF-mass is a constant oF motion; i.e. , the Y cm accelerates downward because the P y compo- nent oF the net momentum is not conserved. Constant X cm means Δ X cm = 0 and there- Fore m Δ X block + M Δ X wedge = 0 . Note that this Formula does not depend on where the wedge has its own center-oF-mass; as long as the wedge is rigid, its overall dis- placement Δ X wedge is all we need to know.
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/28/2008 for the course PHY 301 taught by Professor Swinney during the Spring '07 term at University of Texas at Austin.

### Page1 / 8

hjm - Version 108/ABCDA final Demkov (59910) This print-out...

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

View Full Document
Ask a homework question - tutors are online