Coefficient Of Kinetic Friction Question

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nintendude794
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Homework Statement


A 25 kg box slides down a 26 degree ramp with an acceleration of 1.32 m/s2.
The acceleration of gravity is 9.81 m/s2
Find the coefficient of kinetic friction between the box and the ramp.

Homework Equations


mass = m = 25 kg
angle = (theta) = 26 degrees
acceleration = a = 1.32 m/s2
gravity = Fg = 9.81 m/s2
coefficient of kinetic friction = Mk = Fk/Fn
Fk = resistance, friction? Force opposite Mass*Acceleration.
Fn = perpendicular to the object's motion, upward

The Attempt at a Solution


Fµ = Fn - COS26(I don't know...)

Attached is a pic of my teacher's process for solving the same problem, only with different values.

I'm clueless. Teacher said this is the toughest of the 14 problems that are due in 6 hours, so I thought I might seek assistance in learning how to do it. Please and thanks. :)
 

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The force of kinetic friction is proportional to the normal force Fn, and its acts along the slope in opposite direction as the body moves. Fn, the normal force is the component of mg which is perpendicular to the slope,

Fn=mgcos(theta).

The force of kinetic friction is

Fk=μ mg cos(theta).

The component of mg which is parallel with the slope is

Fp=mg sin(theta),

and it points down the slope. The force of kinetic friction points in the opposite direction. The resultant of these forces will accelerate the body down the slope

ma=mg sin(theta)-μ mg cos(theta).

Given m, a, theta, g, find μ.

ehild
 
ehild said:
The force of kinetic friction is

Fk=μ mg cos(theta).

ehild

Here, what exactly is "μ"? What does it mean?
 
ehild said:
It is the coefficient of friction. You used the notation Mk.

ehild

Thanks. This assignment was due 30 minutes ago, I accepted a zero if even only temporarily because I do not understand. I have printed it out to take it to tutorials. Thanks so much for trying to help me though. :)