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Coefficient of Friction Problems (Beginner Physics)

by rwishka
Tags: beginner, coefficient, friction, physics
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rwishka
#1
Feb2-07, 06:44 AM
P: 16
1. The problem statement, all variables and given/known data

1. A couch with a mass of 1x10^2 kg is placed on an adjustable ramp connected to a truck. As one end of the ramp is raised, the couch begins to move downward. If the couch slides down the ramp with an acceleration of 0.70 m/s^2 when the ramp angle is 25 degrees, what is the coefficient of kinetic friction between the ramp and couch? (g = 9.81 m/s^2)

A. 0.47
B. 0.42
C. 0.39
D. 0.12

2. A crate is carried in a pickup truck traveling horizontally at 15.0 m/s. The truck applies the brakes for distance at 28.7 m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed, if the crate does not slide?

A. 0.400
B. 0.365
C. 0.892
D. 0.656

2. Relevant equations

1. F = ma
u(kinetic) = Force(kinetic)/Force(normal) = Force(kinetic)/(mg)
2. Either v(final) = v(initial) + a(t)
OR
distance = 0.5(initial velocity + final velocity)(t)
OR
distance = v(inital)(t) + 0.5a(t)^2

-I think those are the three formulas I have for constant uniform acceleration. How do I choose which one to use? And what to do after finding the acceleration?

3. The attempt at a solution

1. F=ma
F = (1 x 10^2 kg)(0.70 m/s^2)
F = 70 N

Is this right? How I involve the angle?

2. vf = vi + at
-15 = 430.5a
a = -0.0348

How can you find normal force? I know it's (mg), but is there another way? I bad at drawing diagrams, and working it out.

Please help!
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ritwik06
#2
Feb2-07, 07:49 AM
P: 582
Quote Quote by rwishka View Post
1. The problem statement, all variables and given/known data

1. A couch with a mass of 1x10^2 kg is placed on an adjustable ramp connected to a truck. As one end of the ramp is raised, the couch begins to move downward. If the couch slides down the ramp with an acceleration of 0.70 m/s^2 when the ramp angle is 25 degrees, what is the coefficient of kinetic friction between the ramp and couch? (g = 9.81 m/s^2)

A. 0.47
B. 0.42
C. 0.39
D. 0.12

2. A crate is carried in a pickup truck traveling horizontally at 15.0 m/s. The truck applies the brakes for distance at 28.7 m while stopping with uniform acceleration. What is the coefficient of static friction between the crate and the truck bed, if the crate does not slide?

A. 0.400
B. 0.365
C. 0.892
D. 0.656

2. Relevant equations

1. F = ma
u(kinetic) = Force(kinetic)/Force(normal) = Force(kinetic)/(mg)
2. Either v(final) = v(initial) + a(t)
OR
distance = 0.5(initial velocity + final velocity)(t)
OR
distance = v(inital)(t) + 0.5a(t)^2

-I think those are the three formulas I have for constant uniform acceleration. How do I choose which one to use? And what to do after finding the acceleration?

3. The attempt at a solution

1. F=ma
F = (1 x 10^2 kg)(0.70 m/s^2)
F = 70 N

Is this right? How I involve the angle?

2. vf = vi + at
-15 = 430.5a
a = -0.0348

How can you find normal force? I know it's (mg), but is there another way? I bad at drawing diagrams, and working it out.

Please help!
1. If the object is on a tilted surface such as an inclined plane, the normal force is less, because less of the force of gravity is perpendicular to the face of the plane. Therefore, the normal force, and ultimately the frictional force, is determined using vector analysis, usually via a free body diagram. Depending on the situation, the calculation of the normal force may include forces other than gravity.

The actual reaction will be the weight*sin 25

2. Use the formula:
v^2=(u^2)-2aS
The kinetic force of friction is the force required to keep the body in uniform motion.
AngeloG
#3
Feb2-07, 01:54 PM
P: 103
Draw a picture first! It's good to set up all the forces that are acting on the object. Then to use a free body force diagram to see how they act in the X and Y axis.

You'll have the force of gravity in the X and Y direction; also, your acceleration will set the positives/negatives on the system. Your force of friction should be negative and it should have a component since the force of friction depends on the normal force acting on it.

Also, you will be calling your X at an angle like this / instead of ___ and y \ instead of |. So your axis's will look like X's. One will be X and one will be Y =). In the end there will be acceleration in the x direction, but there will be no acceleration in the Y direction because the object doesn't move up or down while on the x axis.

You'll have 2 sum of forces. Sum of All Forces = mass * acceleration.

{Fx = Forces = MA. From here you will be able to find the coeffecient of friction.
{Fy = Forces = MA (since A = 0) Force = 0. From here you can find your normal.

I'm doing this from a computer lab; but, when I arrive home I'll scan an example so you can follow or learn from it =).


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