Question

physics 7 - unit forces
friction
name lois escobar date 9/19/25
grade & element 7p4

materials μs μk
steel on steel 0.74 0.57
aluminum on steel 0.61 0.47
copper on steel 0.53 0.36
rubber on concrete (dry) 1.0 0.8
rubber on concrete (wet) 0.3 0.25
wood on wood 0.25 - 0.5 0.2
glass on glass 0.94 0.4
teflon on teflon 0.04 0.04
teflon on steel 0.04 0.04
waxed wood on wet snow 0.14 0.1
waxed wood on dry snow 0.10 0.04
metal on metal (lubricated) 0.15 0.06
ice on ice 0.1 0.03
synovial joints in humans 0.01 0.003
very rough surfaces 1.5
calculate the kinetic and static friction for each of the following, the number of newtons in each question represents the weight of the object and the weight is equal to the normal force for all of these questions.

1. a 600 n force on waxed wood skis on dry snow.

1. a 600 n force on waxed wood skis on wet snow.

1. a 9,000 n car with rubber tires on dry concrete.

1. a 9,000 n car with rubber tires on wet concrete.

1. a synovial joint receiving 18 n of pressure.

1. 1.2 newton aluminum can on a steel plate.

Explanation:

Step1: Recall friction - force formula

The formula for static friction is $F_{s}=\mu_{s}N$ and for kinetic friction is $F_{k}=\mu_{k}N$, where $\mu_{s}$ and $\mu_{k}$ are the coefficients of static and kinetic friction respectively, and $N$ is the normal force.

Step2: Solve for problem 1

From the table, for waxed - wood on dry snow, $\mu_{s}=0.10$ and $\mu_{k}=0.04$, and $N = 600$ N.
$F_{s}=\mu_{s}N=0.10\times600 = 60$ N
$F_{k}=\mu_{k}N=0.04\times600 = 24$ N

Step3: Solve for problem 2

For waxed - wood on wet snow, $\mu_{s}=0.14$ and $\mu_{k}=0.1$ from the table, and $N = 600$ N.
$F_{s}=\mu_{s}N=0.14\times600 = 84$ N
$F_{k}=\mu_{k}N=0.1\times600 = 60$ N

Step4: Solve for problem 3

For rubber on dry concrete, $\mu_{s}=1.0$ and $\mu_{k}=0.8$, and $N = 9000$ N.
$F_{s}=\mu_{s}N=1.0\times9000 = 9000$ N
$F_{k}=\mu_{k}N=0.8\times9000 = 7200$ N

Step5: Solve for problem 4

For rubber on wet concrete, $\mu_{s}=0.3$ and $\mu_{k}=0.25$, and $N = 9000$ N.
$F_{s}=\mu_{s}N=0.3\times9000 = 2700$ N
$F_{k}=\mu_{k}N=0.25\times9000 = 2250$ N

Step6: Solve for problem 5

For synovial joints in humans, $\mu_{s}=0.01$ and $\mu_{k}=0.003$, and $N = 18$ N.
$F_{s}=\mu_{s}N=0.01\times18 = 0.18$ N
$F_{k}=\mu_{k}N=0.003\times18 = 0.054$ N

Step7: Solve for problem 6

For aluminum on steel, $\mu_{s}=0.61$ and $\mu_{k}=0.47$, and $N = 1.2$ N.
$F_{s}=\mu_{s}N=0.61\times1.2 = 0.732$ N
$F_{k}=\mu_{k}N=0.47\times1.2 = 0.564$ N

Answer:

1. Static friction: 60 N, Kinetic friction: 24 N
2. Static friction: 84 N, Kinetic friction: 60 N
3. Static friction: 9000 N, Kinetic friction: 7200 N
4. Static friction: 2700 N, Kinetic friction: 2250 N
5. Static friction: 0.18 N, Kinetic friction: 0.054 N
6. Static friction: 0.732 N, Kinetic friction: 0.564 N