1. Friction & Net Force Practice

2. What causes friction?

3. Friction is caused by irregularities between surfaces.

4. How does motor oil in car’s engine help reduce friction?

5. The oil fills in all of the grooves to produce smoother surfaces.

6. The Coefficient of Friction
What is the Greek Symbol that we use to represent the coefficient?
What is the formula for calculating the coefficient?
What are the units for the coefficient?

7. Put these in order from greatest coefficient of static friction to least.
Ice on ice
Rubber on concrete
Wood on brick
Synovial joints of humans
Steel on steel

8. Put these in order from greatest coefficient of static friction to least.
4 Ice on ice (0.1)
1 Rubber on concrete (1.0)
3 Wood on brick (0.6)
5 Synovial joints of humans (0.01)
2 Steel on steel (0.74)

9. Neil Evans pushes his physics ramp across the classroom floor
Neil Evans pushes his physics ramp across the classroom floor. The ramp has a mass of 4.0 kg. The coefficient of friction between the ramp and the floor is How much force must he apply to be able to overcome the friction?

10. 39 N Ffrict = µ Fnorm Ffrict = 0.75 (39N) Ffrict = 29 N 39 N 29 N
4.0 kg
39 N

11. 39 N
Neil Evans is able to push with enough force to overcome friction and then some. He pushes with a force of 50N. With what acceleration will the ramp move?
29 N
4.0 kg
39 N

12. 39 N
Neil Evans is able to push with enough force to overcome friction and then some. He pushes with a force of 50. N. With what acceleration will the ramp move?
Net Force = 50 N –29 = 21 N
F = ma
21 N = (4 kg) a
A = 5.3 m/s2
29 N
4.0 kg
39 N

13. WHO’S READY FOR A CHALLENGE?

14. #1

15. #1 ANSWER Fgrav = m • g = (10 kg) • (9.8 m/s/s) = 98 N
Using the sine function,
Fy = (60 N) • sine (30 degrees) = 30 N
Since vertical forces are balanced, Fnorm = 68 N.
Now Ffrict can be found
Ffrict = mu • Fnorm = ( 0.3) • (68 N) = 20.4 N
Using the cosine function,
Fx = (60 N) • cosine (30 degrees) = 52.0 N
Now since all the individual force values are known, the Fnet can be found:
Fnet = 52.0 N,right N, left = 31.6 N, right.
The acceleration is
a = Fnet / m = (31.6 N) / ( 10 kg)
a = 3.16 m/s/ s, right

16. #2
A 50-N applied force (30 degrees to the horizontal) accelerates a box across a horizontal sheet of ice (see diagram). Glen Brook, Olive N. Glenveau, and Warren Peace are discussing the problem. Glen suggests that the normal force is 50 N; Olive suggests that the normal force in the diagram is 75 N; and Warren suggests that the normal force is 100 N. While all three answers may seem reasonable, only one is correct. Indicate which two answers are wrong and explain why they are wrong.

17. #2 ANSWER
Glen Brook and Warren Peace are incorrect. Warren Peace perhaps believes that the Fnorm = Fgrav; but this is only the case when there are only two vertical forces and no vertical acceleration; sorry Warren - there is a second vertical force in this problem (Fapp).
Glen Brook perhaps thinks that the Fapp force is 50 N upwards and thus the Fnorm must be 50 N upwards to balance the Fgrav. Sorry Glen - the Fapp is only 25 N upwards (50 N) • sine 30 degrees).
"Datagal Olive!"

18. #3
A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform an extensive analysis of the diagram below to determine the values for the blanks.

19. #3 ANSWER First use the mass to determine the force of gravity.
Fgrav = m • g = (20 kg) • (9.8 m/s/s) = 196 N
Now find the vertical component of the applied force using a trigonometric function.
Fy = (80 N) • sine (45 degrees) = 56.7 N
Thus, Fnorm =139.3 N in order for the vertical forces to balance.
The horizontal component of the applied force can be found as
Fx = (80 N) • cosine (45 degrees) = 56.7 N
Since the speed is constant, the horizontal forces must also balance; and so Ffrict = 56.7 N.
The value of "mu" can be found using the equation
"mu"= Ffrict / Fnorm = (56.7 N) / (139.3 N) =

20. #4

21. #4 ANSWER The Fgrav can be calculated from the mass of the object.
Fgrav = m • g = (20 kg) • (9.8 m/s/s) = 196 N
The vertical component of the applied force can be calculated using a trigonometric function:
Fy = (80 N) • sine (30 degrees) = 40 N
In order for the vertical forces to balance, Fnorm + Fy = Fgrav. Thus,
Fnorm = Fgrav - Fy = = 196 N - 40 N = 156 N
The horizontal component of the applied force can be calculated using a trigonometric function:
Fx = (80 N) • cosine (30 degrees) = 69.2 N
The net force is the sum of all the forces when added as vectors. Thus,
Fnet = (69.2 N, right) + (40 N,left) = 29.2 N, right
The acceleration is
a = Fnet / m = (29.2 N, right) / (20 kg) = 1.46 m/s/s, right.
The value of "mu" can be found using the equation "mu" = Ffrict / Fnorm.
"mu" = Ffrict / Fnorm = (40 N) / (156 N) = 0.256

22. #5
Study the diagram below and determine the acceleration of the box and its velocity after being pulled by the applied force for 2.0 seconds.

23. #5 ANSWER The Fgrav can be calculated from the mass of the object.
Fgrav = m • g = (20 kg) • (9.8 m/s/s) = 196 N
The vertical component of the applied force can be calculated using a trigonometric function.
Fy = (50 N) • sine (45 degrees) = 35.4 N
In order for the vertical forces to balance, Fnorm + Fy = Fgrav. Algebraic rearrangement leads to:
Fnorm = Fgrav - Fy = 196 N N = N
The horizontal component of the applied force can be calculated using a trigonometric function:
Fx = (50 N) • cosine (45 degrees) = 35.4 N
The Fnet is 35.4 N, right since the only force which is not balanced is Fx.
The acceleration is:
a = Fnet / m = (35.4 N) / (20 kg) = 1.77 m/s/s
The velocity after 2.0 seconds can be calculated using a kinematic equation:
vf = vi + a • t = 0 m/s + (1.77 m/s/s) • (2.0 s)
vf = 3.54 m/s

24. #6
The following object is moving to the right and encountering the following forces. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object.

25. #6 ANSWER The Fgrav can be calculated from the mass of the object.
Fgrav = m • g = (10 kg) • (10 m/s/s) = 100 N
The vertical component of the applied force can be calculated using a trigonometric function.
Fy = (50 N) • sine (45 degrees) = 35.4 N
In order for the vertical forces to balance, Fnorm + Fy = Fgrav. Algebraic rearrangement leads to:
Fnorm = Fgrav - Fy = 100 N N = 64.6 N
The horizontal component of the applied force can be calculated using a trigonometric function:
Fx = (50 N) • cosine (45 degrees) = 35.4 N
This x-component of the applied force (Fx) is directed leftward. This horizontal component of force is not counteracted by a rightward force. For this reason, the net force is 35.4 N.  Knowing Fnet, allows us to determine the acceleration of the object:
a = Fnet / m = (35.4 N) / (10 kg) = ~3.5 m/s/s
The acceleration of an object is the velocity change per time. For an acceleration of 3.5 m/s/s, the velocity change should be 3.5 m/s for each second of time change. In the velocity-time table, the velocity is decreasing by 3.5 m/s each second. Thus, the values should read 17.5 m/s, 14.0 m/s, 10.5 m/s, 7.0 m/s, 3.5 m/s, 0 m/s.