Friction on an inclined plane

  • Thread starter Cooojan
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  • #1
Cooojan

Homework Statement



There is a mass on incline surface that is acted by foces.

20171024_195755.jpg


μ=0.8
m=2 kg
θ=30°
g=10 m/s^2

##F_N##= normal force
##F_F##= friction force
##Fg_x## = horizontal gravitational component
##Fg_y## = vertical gravitational component

Homework Equations



##F_N=mg~cosθ##
##F_F=μmg~cosθ##
##Fg_x=mg~sinθ##
##Fg_y=mg~cosθ##

The Attempt at a Solution



##∑F_y=F_N-Fg_y=mg~cosθ-mg~cosθ=0##
##∑F_x=Fg_x-F_F=mg~sinθ-μmg~cosθ=mg(sinθ-μ~cosθ)=20(\frac12 -0.8 \frac{√3}{2})=20(-0.193)=-3.86~N##

##a= \frac Fm = -1.93~m/s^2##

This makes absolutely no sence to me.
How can the block have the upward acceleration on the incline... And how can the Friction force be bigger the ##Fg_x##?
 

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Answers and Replies

  • #2
haruspex
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how can the Friction force be bigger the ##Fg_x##?
It can't be, of course.
Can you correctly quote the equation relating normal force, frictional force and coefficient of static friction?
 
  • #3
kuruman
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How can the block have the upward acceleration on the incline
If the block is sliding downhill friction will oppose the motion and be uphill. What is the full statement of the problem? Is the block sliding uphill or downhill? There is nothing wrong about the frictional force being larger than mg sinθ. Think about it.
 
  • #4
Cooojan
If the block is sliding downhill friction will oppose the motion and be uphill. What is the full statement of the problem? Is the block sliding uphill or downhill? There is nothing wrong about the frictional force being larger than mg sinθ. Think about it.

My point is, when we put the 2 kg block on the incline, it cant just start sliding up due to the gravity...
 
  • #5
kuruman
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My point is, when we put the 2 kg block on the incline, it cant just start sliding up due to the gravity...
I agree. That is why we need the full description of the problem. The block could doing any number of things, it could be (a) at rest or moving at constant velocity; (b) sliding downhill and speeding up; (c) sliding downhill and slowing down; (d) sliding uphill and slowing down. Your description
There is a mass on incline surface that is acted by foces.
is not helpful to determine which of the above possibilities is the case.
 
  • #6
Cooojan
I agree. That is why we need the full description of the problem. The block could doing any number of things, it could be (a) at rest or moving at constant velocity; (b) sliding downhill and speeding up; (c) sliding downhill and slowing down; (d) sliding uphill and slowing down. Your description

is not helpful to determine which of the above possibilities is the case.

Actually the exercise is totaly different. I just invented this problem to help me understand, how would the block behave with the following values. But lets say the block is initially at rest
 
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  • #7
Cooojan
It can't be, of course.
Can you correctly quote the equation relating normal force, frictional force and coefficient of static friction?

##F_F= μ~F_N##
##F_N = mg~cosθ~~~~~~⇒~~~~~~F_F=μ~mg~cosθ##
 
  • #8
kuruman
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Actually the exercise is totaly different. I just invented this problem to help me understand, how would the block behave with the ollowing values. But lets say the block is initially at rest
OK. So you picked some values and you want to know what will happen if you place the block on the incline. You should understand that the coefficient of kinetic friction is relevant only if the block is sliding. If that's the case, then the force of kinetic friction is fk = μkFN. If you want to know whether the block will start sliding downhill when you place it on the incline, then you need to compare the downhill component of the weight with the maximum force of static friction that the surface can exert. This is fsmax = μs FN where μs is the coefficient of static friction.

Here you found that the downward component of the weight is mg sinθ = 2 kg×10 m/s2×½ = 10 N
Assuming that μs = 0.8, fsmax = 0.8×2 kg×10 m/s2×0.866 = 13.8 N
Since the surface can exert as much as 13.8 N uphill but only 10 N are needed to keep the block in place, the block stays in place. Note that the force of friction here is not larger than the downhill component of the weight. The uphill force of friction is just what is needed to have zero acceleration, namely 10 N.
 
  • #9
haruspex
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##F_F= μ~F_N##
Wrong.
If you stand still on a level surface, is there a horizontal frictional force on your feet from the floor?
 
  • #10
Cooojan
OK. So you picked some values and you want to know what will happen if you place the block on the incline. You should understand that the coefficient of kinetic friction is relevant only if the block is sliding. If that's the case, then the force of kinetic friction is fk = μkFN. If you want to know whether the block will start sliding downhill when you place it on the incline, then you need to compare the downhill component of the weight with the maximum force of static friction that the surface can exert. This is fsmax = μs FN where μs is the coefficient of static friction.

Here you found that the downward component of the weight is mg sinθ = 2 kg×10 m/s2×½ = 10 N
Assuming that μs = 0.8, fsmax = 0.8×2 kg×10 m/s2×0.866 = 13.8 N
Since the surface can exert as much as 13.8 N uphill but only 10 N are needed to keep the block in place, the block stays in place. Note that the force of friction here is not larger than the downhill component of the weight. The uphill force of friction is just what is needed to have zero acceleration, namely 10 N.


OK! I think I get it. So the block won't be sliding downhill, unless summary force in the downhill direction would be larger that 13,8N...
 
  • #11
Cooojan
Wrong.
If you stand still on a level surface, is there a horizontal frictional force on your feet from the floor?

##F_{Fs} ≤μ_sF_N##

I see
 
  • #12
Cooojan
Thank you guys. Realy helped me out there!!
 

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