# Friction on an Inclined Plane: Solving for Acceleration and Forces

• Cooojan
In summary, this conversation discusses a problem involving a block on an inclined surface and the forces acting on it. The equations for normal force, frictional force, and the horizontal and vertical gravitational components are provided. The conversation also addresses the role of static and kinetic friction in determining whether the block will slide or remain at rest.
Cooojan

## Homework Statement

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

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

##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 sense 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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Cooojan said:
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?

Cooojan said:
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.

kuruman said:
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 can't just start sliding up due to the gravity...

Cooojan said:
My point is, when we put the 2 kg block on the incline, it can't 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
Cooojan said:
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.

kuruman said:
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 let's say the block is initially at rest

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haruspex said:
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##

Cooojan said:
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 let's 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.

Cooojan
Cooojan said:
##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?

Cooojan
kuruman said:
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...

haruspex said:
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

Thank you guys. Realy helped me out there!

## 1. What is friction on an inclined plane?

Friction on an inclined plane is the force that resists the motion of an object as it slides down or up an inclined surface. It occurs due to the interaction between the object and the surface it is moving on.

## 2. What factors affect friction on an inclined plane?

The factors that affect friction on an inclined plane include the type of surface, the weight and material of the object, and the angle of inclination. Rougher surfaces and heavier objects will have higher friction, while smoother surfaces and lighter objects will have lower friction. The angle of inclination also plays a role, as steeper angles will result in higher friction.

## 3. How is the force of friction calculated on an inclined plane?

The force of friction on an inclined plane can be calculated using the formula Ff = μmgcosθ, where Ff is the force of friction, μ is the coefficient of friction, m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of inclination.

## 4. How does friction on an inclined plane affect the speed of an object?

Friction on an inclined plane will decrease the speed of an object as it moves down the incline. This is because the force of friction acts in the opposite direction of the motion, slowing down the object's velocity. However, if the angle of inclination is low and the force of friction is relatively small, the object may still move at a significant speed.

## 5. Can friction on an inclined plane ever be eliminated?

It is not possible to completely eliminate friction on an inclined plane. However, friction can be reduced by using smoother surfaces, lighter objects, or decreasing the angle of inclination. In certain situations, such as in a vacuum or with extremely low friction surfaces, the effects of friction can be minimized, but it can never be completely eliminated.

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