1. Exercise Class For College Physics
俞颉翔(Jiexiang Yu)
Office: 2401, East Guanghua Building
Hello everyone. I’m glad to stand here to have this exercise class, because I can practice my spoken English. The content of this lesson is not very much. 6 problems at all. So I will not have a break in the middle of class. Then we can finish class earlier. And we can discuss after class, if there is any question.

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?
First, let’s see a problem in our recent homework. The error rate is quite high.
Perhaps many of you

3. For x direction: For y direction: 31o T Fl 15o W=mg y x
To solve the problem like this, including static systems, the key is the analysis of forces on a particular point. describe each force as components in a x-y coordinate system. For problem a, we choose this point. There are three forces: gravity of the person, force exerted by legs, tension in the rope.
It is important to understand problem correctly.

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

5. Another solution
Since the system remains stationary, the additions of the force vectors is zero.
15o
31o T Fl
W=mg
75o
74o
Three vectors can build up a triangle.

6. Problem 4.58 on P113 Coefficient of friction f 15o N Fl
The coefficient of friction is a property of the interface of two materials.
It’s independent on the supportive force and the friction.

7. Something about friction
Coefficient of static friction, μs
Coefficient of kinetic friction,μk.
Usually, μs > μk in the same situation
Rolling friction
Sliding friction
Did Mr. Le tell you about this?

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

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

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?
This problem is nothing to do with the friction. However, it’s a dynamic system. m1 m2

11. Since the pulley is mass-less, the net force exerted on it is zero
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: F T T m2
W2 = m2g T

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

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.
this problem is also a dynamic problem. I think it’s more complex than the previous one.
In this problem, we need to consider the friction as well as the acceleration. M m F
No friction

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

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

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?
Then, let’s review the kinematics. It’s nothing to do with forces, so the problem seems to be easy.
21.8m
42.0o
25.3m/s

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:
21.8m
42.0o
25.3m/s

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.
In my opinion, it’s important to identify a positive direction for the displacement, velocity and acceleration. It is helpful for you when a complex calculations is demanded.
21.8m
42.0o
25.3m/s

19. Problem 5
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 adjacent links,
The force F exerted on the top link by the agent lifting the chain,
The net force on each link. 1 2 3 4 5 F
It’s the last problem.

20. For link 1: For link 2: For link 3: For link 4: For link 5: F 5 4 3 2
The key to solve this problem is the acceleration is a conserved quantity in each links.

21. The net force on 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. F x
Although you needn’t to consider this problem as homework, you’d better to think it after class.