Coefficient of static friction on an inclined plane

In summary, the truck loaded with rock has a coefficient of static friction of 0.466 between the rock and steel tray.
  • #1
hoges
2
0

Homework Statement



A dump truck loaded with rock has a coefficient of static friction of 0.4, and a coefficient of kinetic friction of 0.2 between the rock and steel tray. If the truck tilts its bed to 25°, calculate the coefficient of static friction on the load. Hence calculate the angle at which the load begins to slide. The mass of the rock in the truck is 77110kg.

Homework Equations



Ff = mgsinθ
FN = mgcosθ
Ff = μFN

The Attempt at a Solution



Ff = μFN

so

mgsinθ = μmgcosθ

Cancelling mg from both sides gives

sinθ = μcosθ

μ = (sinθ)/(cosθ)

μ = tanθ

μ = tan 25°

μ = 0.466

How can this be? How can μ on an inclined plane be greater than μ between the two flat surfaces? Is this saying that the coefficient of friction on the inclined plane is 0.466 x μ given in the question? Which would give a value of μ=0.1864 for the load on the inclined plane?


Taking this value for μ as 0.466 and using it to determine the angle at which the load begins to slide gives us:

The load will begin to slide when tanθ > 0.466

But this gives θ > 25°

So the load will begin to slide when the tray is tilted to more than 25°?

I have been through every method I can think of and every time I keep coming back to μ = 0.466 and θ > 25° but I am sure that these values are incorrect.
 
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  • #2
Problem is incorrectly worded. It is given that the static friction coefficient is 0.4, so how can it be calculated if it is given? The rocks will start to slide when tan theta = 0.4 or about 22 degrees. At 25 degrees, they will be sliding.
 
  • #3
That's exactly what I did in the first place, figured out at what angle tanθ > 0.4, but then I figured it must have been incorrect since it asked me to determine μ.

Will email the lecturer and clarify what exactly they want us to find, as the problem seems to simple if this is the case.

Thanks for your help.
 

Related to Coefficient of static friction on an inclined plane

1. What is the coefficient of static friction on an inclined plane?

The coefficient of static friction on an inclined plane is a measure of the force required to prevent an object from sliding down the plane. It is represented by the symbol µs and is a dimensionless quantity.

2. How is the coefficient of static friction on an inclined plane calculated?

The coefficient of static friction on an inclined plane can be calculated by dividing the maximum frictional force that can be applied to an object without it sliding, by the normal force acting on the object. The normal force is equal to the weight of the object multiplied by the cosine of the angle of inclination of the plane.

3. What factors affect the coefficient of static friction on an inclined plane?

The coefficient of static friction on an inclined plane is affected by the nature of the surfaces in contact, the angle of inclination of the plane, and the weight of the object. Rougher surfaces generally have higher coefficients of static friction, and as the angle of inclination increases, the coefficient of static friction decreases.

4. How does the coefficient of static friction on an inclined plane affect the motion of an object?

The coefficient of static friction on an inclined plane determines the maximum angle at which an object can remain at rest without sliding down the plane. If the angle of inclination is greater than this maximum angle, the object will begin to slide down the plane due to the force of gravity.

5. What are some real-life applications of the coefficient of static friction on an inclined plane?

The coefficient of static friction on an inclined plane is important in many industries, such as construction and transportation, where objects may need to be moved or secured on inclined surfaces. It is also relevant in sports and recreation, such as rock climbing and skiing, where understanding the coefficient of static friction can help individuals navigate inclined surfaces safely and effectively.

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