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Friction: Basic questions

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1 Observe and Find a Pattern
Examine the data in the table that follows:
Mass of the block Surface area Quality of surfaces Maximum static friction force
1 kg 0.1 m2 Medium smooth 3.1 N
1 kg 0.2 m2 Medium smooth 3.0 N
1 kg 0.3 m2 Medium smooth 3.1 N
1 kg 0.1 m2 A little rougher 4.2 N
1 kg 0.1 m2 Even rougher 5.1 N
1 kg 0.1 m2 Roughest 7.0 N

a) Now decide how the maximum static friction force that the surface exerts on the block depends on the surface area of the block and on the roughness of the two surfaces.

3 Observe and Find a Pattern
Examine the data in the table that follows.
Mass of the block Extra downward force exerted on the 1-kg block Normal force exerted by the board on the block Maximum
static friction force
1 kg 0 N 10 N 3 N
1 kg 5 N 15 N 4.5 N
1 kg 10 N 20 N 6 N
1 kg 20 N 30 N 9 N

a) Use the data in the table to find the relationship between the maximum static friction force and the normal force the surface exerts on the block.
b) Express mathematically a relationship between the normal force and the maximum static friction force. Write this relationship as an equation.

4 Represent and Reason
Some students are trying to move a GIANT pumpkin across the room. Angelique pulls it across the floor at the same time that Gunnar and Jalen pull on it from the other side. Gunnar pulls on the pumpkin, exerting a (-150-N) force, and Jalen pulls exerting a (-125-N) force. There is also a (-200-N) friction force exerted by the floor on the pumpkin. The sum of the forces exerted on the desk is (+27-N).
a) Make a sketch of the situation.
b) Draw a force diagram for the pumpkin. Draw a motion diagram.
c) Write an algebraic statement that describes the force diagram you drew.
d) How hard is Angelique pulling?
e) Is the pumpkin moving with a constant velocity or is it speeding up? How do you know?

5 Represent and Reason
A 50-kg box of candy rests on the floor. The coefficients of static and kinetic friction between the bottom of the box and the floor are 0.70 and 0.50, respectively.
a) What is the minimum force a person needs to exert on it to start the box sliding?
b) After the box starts sliding, the person continues to push it exerting the same force. What is the acceleration of the box?

9 Represent and Reason Olympic skier Lindsey Vonn skis down a steep slope that descends at an angle of 30O below the horizontal. The coefficient of sliding friction between her skis and the snow is 0.10. Determine Vonn's acceleration, and her speed 6.0s after starting.

10 Represent and Reason
Bode Miller, 80-kg downhill skier, descends a slope inclined at 20O. Determine his acceleration if the coefficient of friction is 0.10. How would this acceleration compare to that of a 160-kg skier going down the same hill?

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SOLUTION This solution is FREE courtesy of BrainMass!

Friction (willing to add credits if necessary)
1 Observe and Find a Pattern
Examine the data in the table that follows:
Mass of the block Surface area Quality of surfaces Maximum static friction force
1 kg 0.1 m2 Medium smooth 3.1 N
1 kg 0.2 m2 Medium smooth 3.0 N
1 kg 0.3 m2 Medium smooth 3.1 N
1 kg 0.1 m2 A little rougher 4.2 N
1 kg 0.1 m2 Even rougher 5.1 N
1 kg 0.1 m2 Roughest 7.0 N
a) Now decide how the maximum static friction force that the surface exerts on the block depends on the surface area of the block and on the roughness of the two surfaces.
The maximum static friction force (limiting friction) is nearly remains unchanged with the change in area of contact.
3 Observe and Find a Pattern
Examine the data in the table that follows.
Mass of the block Extra downward force exerted on the 1-kg block Normal force exerted by the board on the block Maximum
static friction force
1 kg 0 N 10 N 3 N
1 kg 5 N 15 N 4.5 N
1 kg 10 N 20 N 6 N
1 kg 20 N 30 N 9 N
Use the data in the table to find the relationship between the maximum static friction force and the normal force the surface exerts on the block.
The maximum friction force increases with the normal reaction of the board.
b) Express mathematically a relationship between the normal force and the maximum static friction force. Write this relationship as an equation.
As the normal reaction on the board increases the maximum static friction force is increases in the same ration. Hence we can say that the maximum static friction force is directly proportional to the normal reaction of the board exerted on the block. Thus we can write
F_max ∝N
Or F_max=costant*N (the constant here is called coefficient of friction)
4 Represent and Reason
Some students are trying to move a GIANT pumpkin across the room. Angelique pulls it across the floor at the same time that Gunnar and Jalen pull on it from the other side. Gunnar pulls on the pumpkin, exerting a (-150-N) force, and Jalen pulls exerting a (-125-N) force. There is also a (-200-N) friction force exerted by the floor on the pumpkin. The sum of the forces exerted on the desk is (+27-N).
a) Make a sketch of the situation.
(I think you can draw it using pencil.)

Draw a force diagram for the pumpkin. Draw a motion diagram.
The forces acting on the pumpkin are
The force exerted by Angelique FA =?
The force exerted by Gunnar FG = - 150 N
The foece exerted by Jalen FJ = - 125 N
Friction force Ffr = -200 N
weight of the pumpkin (vertically downwards) W
Normal reaction of the floor (vertically upwards) N

c) Write an algebraic statement that describes the force diagram you drew.
The pumpkin is not getting lifted from the floor that means that net vertical force on it is zero. Hence the normal reaction must be equal and opposite to the weight of the pumpkin.
The net horizontal force is the vector sum of all the forces acting on it thus
W+N=0
And F_net= F_A+F_G+F_J+F_fr
d) How hard is Angelique pulling?
F_net= F_A+F_G+F_J+F_fr
As we know that the sum of the forces on the pumpkin is + 27 N, substituting the numerical values in the equation above we get
+27N= F_A+(-150N)+(-125N)+(-200 N)
Or F_A=+27-150-125-200=502 N
Hence Angelique is pulling it with a force of 502 N.

e) Is the pumpkin moving with a constant velocity or is it speeding up? How do you know?
It is moving with a speeding velocity.
As the sum of all the forces is not zero according to Newton's law it will be accelerated means it will move with increasing velocity.
5 Represent and Reason
A 50-kg box of candy rests on the floor. The coefficients of static and kinetic friction between the bottom of the box and the floor are 0.70 and 0.50, respectively.
a) What is the minimum force a person needs to exert on it to start the box sliding?
The normal reaction of the floor on the box is equal to its weight and upward thus
N = mg = 50*9.8 = 490 N
Thus the limiting (maximum static) friction force on the box will be
F_max= -μ_s*N
or F_max=- 0.70*490=- 343 N
Hence minimum force required to slide the box is slightly greater than 343 N.
After the box starts sliding, the person continues to push it exerting the same force. What is the acceleration of the box?
The kinetic friction on the box when it start sliding is given by
F_k= -μ_k*N
or F_k=- 0.50*490=- 245 N
Thus the net force acting on the box when it is sliding will be given by
F=343-245=98 N
Thus from Newton's second law of motion acceleration of the block is given by
a=F/m=98/50=1.96 m/s^2
Hence acceleration of the block will be 1.96 ms-2
9 Represent and Reason Olympic skier Lindsey Vonn skis down a steep slope that descends at an angle of 30O below the horizontal. The coefficient of sliding friction between her skis and the snow is 0.10. Determine Vonn's acceleration, and her speed 6.0s after starting.

The forces acting on the skier are
Its weight W = mg (vertically downwards)
Normal reaction N of the slope (normal to the slop)
Kinetic Friction = N along the slope opposing motion.
Resolving the weight of the skier along and normal to slope we get
Component of the weight along the slop will be W sin  ( = 300) and the component normal to the slope will be W cos .
As the skier is not moving in the direction normal to the slope, normal forces must be balanced and thus
N-W cos⁡〖θ 〗=0
Gives N=W cos⁡〖θ 〗 ------------------------------ (1)
The sum of the forces along the slope will be
F=W sin⁡θ- μ_k N
Or F=W sin⁡θ- μ_k*W cos⁡〖θ 〗 (using equation 1)
Hence acceleration of the skier along the slope will be
a=F/m=(W sin⁡θ- μ_k*W cos⁡〖θ 〗)/m
Or a=W(sin⁡θ- μ_k*cos⁡〖θ 〗 )/m
Or a=mg(sin⁡θ- μ_k*cos⁡〖θ 〗 )/m
Or a=g(sin⁡θ- μ_k*cos⁡〖θ 〗 )
Substituting numerical values of the quantities we get
a=9.8(sin⁡〖〖30〗^0 〗- 0.10*cos⁡〖〖30〗^0 〗 )
Or a=9.8(0.500- 0.10*0.866)=4.05 m/s^(-2)
Thus the acceleration of the skier is 4.05 ms-2.
Thus her final velocity v after time t = 6.0 s is given by the first equation of motion
v=u+a*t
Or v=0+4.05*6.0=2.43 s
Hence her velocity 6.0 s after start will be 2.4 m/s
10 Represent and Reason
Bode Miller, 80-kg downhill skier, descends a slope inclined at 20O. Determine his acceleration if the coefficient of friction is 0.10. How would this acceleration compare to that of a 160-kg skier going down the same hill?
Using the same method the acceleration of this skier is given by
a=g(sin⁡θ- μ_k*cos⁡〖θ 〗 )
Here  = 200, g and k is the same thus
a=9.8(sin⁡〖〖20〗^0 〗- 0.10*cos⁡〖〖20〗^0 〗 )
Or a=9.8(0.342- 0.10*0.940)=2.43 m/s^(-2)
Thus the acceleration of Bode Miller is 2.43 ms-2.
As the acceleration of the skier does not depend on his mass, the acceleration remains the same 2.43 ms-2.

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