# Mechanics: Find Coefficient of Friction on 34° Slope w/ 0.4

• MHB
• Shah 72
In summary, the particle slides up a slope at an angle of 34 degrees to the horizontal with a coefficient of friction of 0.4. It passes a point on the way up the slope with a speed of 3 meters per second and passes the point on the way down the slope with a speed of 2 meters per second.
Shah 72
MHB
A particle slides up a slope at angle 34 degree to the horizontal with coefficient of friction 0.4. It passes a point P on the way up the slope with 3m/s and passes it on the way down the slope with speed 2 m/s. Find the coefficient of friction between the particle and the slope.
Coefficient of friction=0.4
F=m×a
While going up
-0.4×10cos 34-10sin 34=a
a=-8.91m/s^2.
V^2=u^2+2as
I get the eq
V^2=9-2×8.91s
I don't understand how to calculate further.
Pls help

Shah 72 said:
A particle slides up a slope at angle 34 degree to the horizontal with coefficient of friction 0.4. It passes a point P on the way up the slope with 3m/s and passes it on the way down the slope with speed 2 m/s. Find the coefficient of friction between the particle and the slope.
Coefficient of friction=0.4
F=m×a
While going up
-0.4×10cos 34-10sin 34=a
a=-8.91m/s^2.
V^2=u^2+2as
I get the eq
V^2=9-2×8.91s
I don't understand how to calculate further.
Pls help

wait ... what?

skeeter said:
wait ... what?
I don't know how to find the coefficient of friction as if I take v^2= u^2+ 2as, there are two unknown factors which is v and s. So I don't know how to calculate.

Shah 72 said:
I don't know how to find the coefficient of friction as if I take v^2= u^2+ 2as, there are two unknown factors which is v and s. So I don't know how to calculate.
skeeter is pointing out that you were given the answer in the problem statement. Look at the bold sections.

-Dan

skeeter said:
wait ... what?
I tried but iam getting the same ans. I have no clue.

Shah 72 said:
I tried but iam getting the same ans. I have no clue.

Did you read the problem statement? It clearly states the coefficient of friction is 0.4 … why then are you trying to find the coefficient of friction when it’s given?

skeeter said:
Did you read the problem statement? It clearly states the coefficient of friction is 0.4 … why then are you trying to find the coefficient of friction when it’s given?
The question asks me to. That's why this question is so confusing. And the ans in the textbook is 0.259

Shah 72 said:
The question asks me to. That's why this question is so confusing. And the ans in the textbook is 0.259
Then something is wrong with the problem statement. Talk to your instructor.

-Dan

up the incline …

$a = -[10\sin(34)+4\cos(34)] = -8.91 \, m/s^2$

$\Delta x = \dfrac{0^2-3^2}{2a} = 0.51 \, m$

down the incline …

$a = \dfrac{2^2-0^2}{2 \Delta x} = 3.96 \, m/s^2$

$\mu = \dfrac{10\sin(34) - a}{10\cos(34)} = 0.2$

not a very well engineered problem, imo.

skeeter said:
up the incline …

$a = -[10\sin(34)+4\cos(34)] = -8.91 \, m/s^2$

$\Delta x = \dfrac{0^2-3^2}{2a} = 0.51 \, m$

down the incline …

$a = \dfrac{2^2-0^2}{2 \Delta x} = 3.96 \, m/s^2$

$\mu = \dfrac{10\sin(34) - a}{10\cos(34)} = 0.2$

not a very well engineered problem, imo.
Tha
skeeter said:
up the incline …

$a = -[10\sin(34)+4\cos(34)] = -8.91 \, m/s^2$

$\Delta x = \dfrac{0^2-3^2}{2a} = 0.51 \, m$

down the incline …

$a = \dfrac{2^2-0^2}{2 \Delta x} = 3.96 \, m/s^2$

$\mu = \dfrac{10\sin(34) - a}{10\cos(34)} = 0.2$

not a very well engineered problem, imo.
Thank you very very much!

## 1. How do you calculate the coefficient of friction on a 34° slope with a value of 0.4?

In order to calculate the coefficient of friction on a 34° slope with a value of 0.4, you will need to use the formula μ = tanθ, where μ represents the coefficient of friction and θ represents the angle of the slope. In this case, μ = tan(34°) = 0.67. Therefore, the coefficient of friction is 0.67.

## 2. What is the significance of the coefficient of friction in mechanics?

The coefficient of friction is a measure of the amount of resistance between two surfaces in contact. In mechanics, it is used to determine the force needed to move an object over a surface, as well as the stability of an object on a slope.

## 3. How does the angle of the slope affect the coefficient of friction?

The angle of the slope has a direct effect on the coefficient of friction. As the angle of the slope increases, the coefficient of friction also increases. This means that it becomes more difficult to move an object on a steeper slope compared to a shallower slope.

## 4. What factors can affect the accuracy of calculating the coefficient of friction?

There are several factors that can affect the accuracy of calculating the coefficient of friction, such as the roughness of the surfaces in contact, the presence of any lubricants or contaminants, and the temperature of the surfaces. These factors can alter the amount of friction between the surfaces and therefore impact the calculated coefficient of friction.

## 5. How can the coefficient of friction be used in real-world applications?

The coefficient of friction is an important factor in many real-world applications, including transportation and engineering. It is used to design and improve the efficiency of vehicles, such as cars and trains, by minimizing the amount of friction between moving parts. It is also used in the construction of buildings and structures to ensure stability and prevent slipping or sliding on slopes.

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