{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HS12 - Adams Russell Homework 12 Due Dec 1 2006 11:00 pm...

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

Adams, Russell – Homework 12 – Due: Dec 1 2006, 11:00 pm – Inst: Opyrchal, H 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A solid sphere of radius R and mass m is held against a wall by a string being pulled at an angle θ . W P F θ R Determine the torque equation about the point P. 1. m g R = F (1 - sin θ ) R 2. m g R = F (1 + sin θ ) R correct 3. m g R = F (1 - cos θ ) R 4. m g R = F (1 + cos θ ) R 5. m g R = F R 6. m g R = 2 F R Explanation: W P F θ R R (1 + sin θ ) A θ The clockwise torque about point P is τ cw = AP × F , where AP = R (1 + sin θ ) and the counter- clockwise torque is τ ccw = m g R . From equilibrium, we obtain the equation τ ccw = τ cw = m g R = R (1 + sin θ ) F , which is our answer. keywords: 002 (part 1 of 1) 10 points The system shown in the figure 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 45 with the horizontal. The coeffi- cient of the static friction between the 15 kg mass and the surface on which it rests is 0 . 38. The acceleration of gravity is 9 . 8 m / s 2 .

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

View Full Document
Adams, Russell – Homework 12 – Due: Dec 1 2006, 11:00 pm – Inst: Opyrchal, H 2 15 kg m 45 What is the largest mass m can have and still preserve the equilibrium? Correct answer: 5 . 7 kg. Explanation: Let : M = 15 kg , m = 5 . 7 kg , and θ = 45 . Basic Concepts: The net force exerted on an object in equilibrium equals zero. Solution: For the system to remain in equi- librium, the net forces on both M and m should be zero; and thus the tension in the rope has an upper bound value T max T max cos θ = μ M g , so (1) T max = μ M g cos θ = (0 . 38) (15 kg) (9 . 8 m / s 2 ) cos(45 ) = 78 . 998 N .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

HS12 - Adams Russell Homework 12 Due Dec 1 2006 11:00 pm...

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

View Full Document
Ask a homework question - tutors are online