Exercise Class For College Physics

1. Exercise ClassFor College Physics 俞颉翔(Jiexiang Yu) 2010-10-20 Email: 09110190010@fudan.edu.cn Office: 2401, East Guanghua Building

2. Problem 4.58 on P113 • Consider the 52kg mountain climber. • Find the tension in the rope (T) and the force that the climber must exert with her feet on the vertical rock face to remain stationary (Fl). • What is the minimum coefficient of friction between her shoes and the cliff?

3. For x direction: • For y direction: 31o y T Fl x 15o W=mg

4. Problem 4.58 on P113——Sine Law R refers to the radius of circumscribed circle in a triangle.

5. Another solution • Since the system remains stationary, the additions of the force vectors is zero. 31o T W=mg 74o 75o Fl 15o

6. Problem 4.58 on P113 • Coefficient of friction f 15o N Fl

7. Something about friction • Coefficient of static friction, μs • Coefficient of kinetic friction,μk. • Usually, μs>μk in the same situation

8. Problem 1 • A horizontal force F = 12N pushes a block weighing 5N against a vertical wall. μs = 0.60 and μk = 0.40. Assume the block is not moving initially. Will the block start moving? F W

9. N = F =12N • The maximum of static friction is: • So the block remains stationary. f F N W W

10. Problem 2 F • Someone exerts a force F directly up on the axle of the pulley. Consider the pulley and string to be mass-less and the bearing frictionless. Two object, m1=1.2kg and m2=1.9kg, are attached to the opposite ends of the string, which passes over the pulley. The m2 is in contact with the floor. • Find the largest value of F may have so that m2 will remain at rest on the floor. • what is the tension in the string if the upward force F is 110N? • With the tension determined in b), what is the acceleration of m1? m1 m2

11. F • Since the pulley is mass-less, the net force exerted on it is zero. Thus the tension in the string: Then we get the maximum of the tension: and the maximum of F: T T T m2 W2 = m2g

12. The tension in the string: • The acceleration of m1: T T m1 m2 W 1= m1g W2=m2g

13. Problem 3 • The two blocks, m = 16kg and M = 88kg are free to move. The coefficient of static friction between the blocks is μs= 0.38, but the surface beneath M is frictionless. Find the minimum horizontal force F required to hold m against M. F M m No friction

14. Acceleration of the two blocks: • Acceleration of the m block: • The friction : • Acceleration of the M block: f N F m W=mg N’ M F N M m f No friction W=Mg No friction

15. From the same acceleration of the two blocks, we have: • Put this into the function of the friction f: f N F m W=mg N’ M F N M m f No friction W=Mg No friction

16. Problem 4 • You throw a ball with a speed of 25.3m/s at an angle of 42.0o above the horizontal directly toward a wall. The wall is 21.8m from the release point of the ball. • how long is the ball in the air before it hits the wall? • how far above the release point does the ball hit the wall? • What are the horizontal and vertical components of its velocity as it hits the wall? • Has it passed the highest point on its trajectory when it hits? 25.3m/s 42.0o 21.8m

17. Solution • The time taken for the ball to hit the wall: • The Vertical distance above the release point as the ball hits the wall: 25.3m/s 42.0o 21.8m

18. Solution • The vector of velocity as it hits the wall: • Since vy > 0, the ball hasn’t passed the peak point of its trajectory. 25.3m/s 42.0o 21.8m

19. Problem5 • A chain consisting of five links, each with mass 100g, is lifted vertically with a constant acceleration of 2.5m/s2. Find: • The forces acting between adjacentlinks, • The force F exerted on the top link by the agent lifting the chain, • The net force on each link. F 5 4 3 2 1

20. For link 1: • For link 2: • For link 3: • For link 4: • For link 5: F 5 4 3 2 1

21. F • The net forceon each link: • Question： • A vertical force F is exerted on a rope of which mass is M and length L with a constant acceleration of a. Find the tension as a function of the distance x from the bottom of the rope. x