# Please check my work (friction on an incline plane)

• iScience
In summary, two blocks of unequal mass are connected by a string over a smooth pulley. One block of mass 2m rests on an incline while the other block of mass m is hung over a pulley on the incline. The coefficient of friction is μk. To find the angle of the incline that allows the blocks to move at a constant speed, the equations mg=(2m)gsinθ+(μk)(2m)gcosθ and (1/2)=(sinθ+(μk)cosθ) were used. However, the equations do not take into account the tension force and friction force. It is possible that the results depend on the direction of the blocks.

#### iScience

two blocks of unequal mass are connected by a string over a smooth pulley. one mass of mass 2m rests on the incline while the second mass of mass m is hung over a pulley over the incline if the coefficient of friction is μk, what angle of the incline allows the masses to move at a constant speed?|F1| = |F2|

F1= weight of mass hanging over the pulley

F2=Ʃforce acting on mass (2m) ie.. the x component of force along the direction of the incline + friction

mg=(2m)gsinθ+(μk)(2m)gcosθ

mg=(2mg)(sinθ+(μk)cosθ)

mg's cancel..

(1/2)=(sinθ+(μk)cosθ)

but since i know that.. μk=tanθ,

1/2=sinθ+tanθcosθ = sinθ+sinθ= 2sinθ

sinθ=(1/4)

θ=14.4775°

is the way i substituted tanθ for the coefficient valid? if not, why is it not valid?

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Why should the force from the hanging mass be equal to the horizontal component of the force on the other mass? I would assume the string goes along the incline.

mg=(2m)gsinθ+(μk)(2m)gcosθ
This looks like the mass m is hanging now, instead of the mass 2m. How did you get that formula?

The result will certainly depend on μk, a specific number cannot be right as result.

Please add explanations to your formulas, otherwise it is hard to understand what you did and what was wrong.

i fixed the m and 2m error.

i set mg equal to the rest of that because the problem wanted me to find the theta when the net force is zero.

so starting here,

mg=(2m)gsinθ+(μk)(2m)gcosθ

factoring mg..

mg=(2mg)(sinθ+(μk)cosθ)

mg's cancel..

(1/2)=(sinθ+(μk)cosθ)

now I'm supposed to solve for theta but i don't know how.. :(

You have to start earlier. Did you draw a sketch? Did you identify all forces acting on the system?
Which forces have to be in equilibrium?

(sinθ)^2+(cosθ)^2=1 might be handy somewhere in the solving process, but first you need correct equations for the forces.

iScience said:
here is the free body diagram with the arbitrary direction-sign specification (although that part wasn't needed)

http://i.imgur.com/BCmluRZ.png
You left out the tension force and the friction.

Hint: If the speed is constant, what must be the tension?

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i essentially set the two tensions equal to one another. where..

T1=mg
T2=(2m)gsinθ+(μk)(2m)gcosθ

does this depend on the direction of the blocks? since if the blocks move to the right..

mg=(2m)gsinθ+(μk)(2m)gcosθ

and if the blocks move to the left..

mg=(2m)gsinθ-(μk)(2m)gcosθ

so this is actually $\hat{v}$ dependent?

Solve it both ways and see if you get reasonable answers.

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

Friction on an inclined plane refers to the force that resists the motion of an object as it moves on an incline. It is caused by the contact between the object and the surface of the incline, and it acts in the opposite direction of the object's motion.

## 2. Why is friction important to consider on an inclined plane?

Friction is important to consider on an inclined plane because it affects the motion and stability of objects on the incline. It can either help or hinder the object's movement, and understanding its effects is crucial in many practical applications, such as designing ramps or calculating the speed of an object on a hill.

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

The force of friction on an inclined plane can be calculated using the formula F = μN, where F is the force of friction, μ is the coefficient of friction, and N is the normal force (perpendicular to the incline) exerted on the object.

## 4. What factors affect the amount of friction on an inclined plane?

The amount of friction on an inclined plane is affected by the weight of the object, the type of surface it is moving on, and the angle of the incline. The coefficient of friction, which is a measure of the roughness of the surface, also plays a significant role in determining the amount of friction.

## 5. How can the effects of friction on an inclined plane be reduced?

The effects of friction on an inclined plane can be reduced by using lubricants on the surface, choosing smoother materials for the incline, or decreasing the angle of the incline. In some cases, adding wheels or ball bearings to the object can also reduce friction and improve its motion on the incline.