Calculate the coefficient of friction from mass and angle

  • #1

Homework Statement


Determine the coefficient of friction for a surface when the mass of an object is 300g and the angle of incline when the object starts to move is 29°
Distance travelled and acceleration are not provided

Homework Equations


W = mg
Ffric = μN
N = mg cos(θ)
μN = mg sin(θ)

The Attempt at a Solution


I started by drawing a free-body diagram and identified the lines of frictional force, normal reaction force and gravity.

I then calculated the weight:
W = mg
= 0.300 x 9.81 m s-2
= 2.94 N

Not sure on the frictional force as μ is not known

N = mg cos(θ)
= 2.94 N x cos(29)
= 2.57 N

This is where I get stuck as my experience in maths and science is fairly basic.

Could someone please talk me through things a bit please?

Thank you.
 

Answers and Replies

  • #2
PeroK
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Homework Statement


Determine the coefficient of friction for a surface when the mass of an object is 300g and the angle of incline when the object starts to move is 29°
Distance travelled and acceleration are not provided

Homework Equations


W = mg
Ffric = μN
N = mg cos(θ)
μN = mg sin(θ)

The Attempt at a Solution


I started by drawing a free-body diagram and identified the lines of frictional force, normal reaction force and gravity.

I then calculated the weight:
W = mg
= 0.300 x 9.81 m s-2
= 2.94 N

Not sure on the frictional force as μ is not known

N = mg cos(θ)
= 2.94 N x cos(29)
= 2.57 N

This is where I get stuck as my experience in maths and science is fairly basic.

Could someone please talk me through things a bit please?

Thank you.
Think about sticking with the maths. Calculating the normal force doesn't seem necessary.
 
  • #3
Thank you for the reply PeroK.

I have calculated the answer to be 0.554.

Not overly confident on it being correct. I divided sin(29) by cos(29).

#clutchingatstraws :(
 
  • #4
PeroK
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Thank you for the reply PeroK.

I have calculated the answer to be 0.554.

Not overly confident on it being correct. I divided sin(29) by cos(29).

#clutchingatstraws :(
Why do you think that's clutching at straws? If you go back to the relevant equations, can you see how to get:

##\mu = \tan \theta##
 
  • #5
As you may have guessed, trigonometry is not a strong point for me. I'm really struggling.

From my basic understanding, opposite divided by adjacent = tan(θ)

That is what I was working with.

I also found an explanation that advised to combine, then simplify the equations:
1) N = mg cos(θ)
2) μN = mg sin(θ)

which ends with μ = \frac sin(θ) cos(θ)

Hence my attempt at answering.

Feel so frustrated and useless.
 
  • #6
haruspex
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Feel so frustrated and useless.
Why? You used the right equations and got the right answer.
That said, let me clarify some...

First, the question omits to specify you are to find the coefficient of static friction. If it wanted kinetic friction it would need to provide data on acceleration.

Secondly, your equation 2) should properly be two equations:

2a) ##\mu_sN≥F_s## for static friction
2b) ##\mu_kN=F_k## for kinetic friction.
Note the ≥. You can assume equality in the current thread because you are told the angle is the point at which it starts to slip, i.e. when the static friction reaches its limit.

Also note that I specify the frictional forces in the equations as Fk/s rather than "mg sin(θ)". That makes the equations completely general. In other situations the force tending to make the mass slide (i.e. the applied force parallel to the surface) might not be mg sin(θ). I would recommend deleting your equation 2) from your notes and replacing it with 2a) and 2b). Otherwise, you will need to remember that the equation 2) only applies in certain contexts. I see many students go wrong because they forget that.

The same problem arises with equation 1). E.g. if some other force were pushing horizontally on the block then the normal force between block and plane would not be mg cos(θ).
 
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