1. Friction and Normal Forces
Physics
Friction and Normal Forces

2. Equations
Fs= (Us)N
F=-k(Delta x)
F=ma
W=mg

3. Practice:
A wooden pallet carrying a load of 600 kg rests on a wooden floor.
A) forklift driver decides to push it without lifting it. What force must be applied to just get the pallet moving?
(if you looked up wood on wood, the coefficient is .28)

4. Solution:
To make the pallet move, the driver must push it with the max static friction
F= fs= Us N = Us (mg)
F= (.28)(600)(9.8)
= 1650 N to move the pallet

5. Part B.
After a bit of time, the pallet begins to slide. How fast is the pallet moving after 0.5 seconds of sliding under the same force you calculated in part a?
(Uk in the chart now equals .17)

6. Solution: Fk = (Uk)(mg) .17(600)(9.8) = 1000N
Forklift is still pushing with 1650
So the net force in the x direction = 1650 – 1000 = 650 N
Use f=ma
650= 600(a)
A= 1.08 m/s^2
Now we use our kinematic equations
Vf=vi + at
Vf= 0 +(1.08)(.5)= .54 m/s

7. Part C.
If the forklift stops pushing, how far does the pallet slide before coming to a stop?

8. Solution: The net force is now the kinetic force (-1000)
Use f=ma again
-1000=600(a) a= m/s2
Then use kinematic equation of
Vf^2 = vi ^2 + 2 a delta x
0^2 = .54 ^2 + 2(-1.67)(delta x) 0=
X=.087 meters

9. A sled of 300 N is moved at a constant speed over a horizonal floor by a force of 50 N parallel to the floor
Determine the coefficient of kinetic friction Us

10. Answer: .166
Fk= UkFn
50 = Uk(300)
Uk= .166

11. Suppose a 1 kg mass is attached to a 4 kg block on the table
If the coefficient of kinetic friction is .20, what is the acceleration of the block?

12. Solution: Fy = -(4kg*9.81) + (4*9.81) =0
Mg + F normal = 0
Fx= (+1kg * 9.81) + (-39.24*.20)=
Mass1* gravity + (F normal for M2 * Uk)
F friction = -7.8
= 1.96
This is our net force
F=ma
1.96=(4)a a= .49 m/s^2

13. A 84 kg skier is gliding down a 20 degree hill with a Uk of .15
What is the frictional force opposing the skier?

14. Solution: Find the complementary angle
= 70 degrees from x axis is where our normal force is applied
f= ma
F= (84)(9.81) = total
Need to take the sin and cos of this value to get the forces in the x and y comonents
824.3 cos 70 = x = 282N
824.3 sin 70 =y= 775 N
Fk = Uk(force normal)
Fk = .15 ( 775 N) = 116 N

15. Now calculate the acceleration of the skier
Take the forces in the x direction and find the net force
282 + (–116) = 166N forward
F=ma
166 N = 84kg(a)
A= 1.97 m/s^s