Determining the angle while dealing with friction on an inclined plane

In summary: Wrong.Okay, I've read through your post carefully and this is what I have to say. By "the first is right if you put the values in the right place," I would think you are implying that the question is correct, but the result cannot be correct since the angle is wrong. For my equation on the friction force, you are right, but the mistake was made when I was putting it in latex. I actually calculated it like this:## F_r = μH * m * g * cos(\alpha) ##Which, after checking, is also wrong.
  • #1
sylent33
39
5
Homework Statement
A box weighing 5kg stands on a board (coefficient of static friction μH = 0.7 and coefficient of sliding friction μG = 0.55). At what angle does the box slide? How big is the acceleration?
Relevant Equations
$$ μH = tan(\alpha) $$
Hello! So the way I have tried to solve this problem is the following;Since it is an inclined plane and the cofficient of static friction is known, getting to the angle at which the box starts sliding is the following

##μH = \frac {sin (\alpha)} {cos(\alpha)} = μH = tan(\alpha) ##

## \alpha = tan(0,7)##

## \alpha = 0,012 (degrees) ##

Now for the acceleration I thought of this;

First I would need to find the force that is pulling the object downhill; I did that like this

## F_d = m * g * sin(\alpha)##

## F = 0,0104 N ##

Than I would have to find friction force, that is "opposing" the downhill force;

## F_r = m * g * cos(\alpha)##
## F = 34,33 N ##

Now the problem here is pretty obvious. According to my calculation the box shouldn't be even moving, it should be standing still. Which has to mean one of three things; 1) My calculation of the angle was wrong 2) My calculation of the forces was wrong 3) My entire interpretation of the problem was wrong.

Would anyone be able to assist me? Kind Regards.
 
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  • #2
sylent33 said:
Homework Statement:: A box weighing 5kg stands on a board (coefficient of static friction μH = 0.7 and coefficient of sliding friction μG = 0.55). At what angle does the box slide? How big is the acceleration?
Relevant Equations:: $$ μH = tan(\alpha) $$

Hello! So the way I have tried to solve this problem is the following;Since it is an inclined plane and the cofficient of static friction is known, getting to the angle at which the box starts sliding is the following

##μH = \frac {sin (\alpha)} {cos(\alpha)} = μH = tan(\alpha) ##

## \alpha = tan(0,7)##

## \alpha = 0,012 (degrees) ##

Now for the acceleration I thought of this;

First I would need to find the force that is pulling the object downhill; I did that like this

## F_d = m * g * sin(\alpha)##

## F = 0,0104 N ##

Than I would have to find friction force, that is "opposing" the downhill force;

## F_r = m * g * cos(\alpha)##
## F = 34,33 N ##

Now the problem here is pretty obvious. According to my calculation the box shouldn't be even moving, it should be standing still. Which has to mean one of three things; 1) My calculation of the angle was wrong 2) My calculation of the forces was wrong 3) My entire interpretation of the problem was wrong.

Would anyone be able to assist me?Kind Regards.
##μH = \frac {sin (\alpha)} {cos(\alpha)} = μH = tan(\alpha) ##

## \alpha = tan(0,7)##

## \alpha = 0,012 (degrees) ##

The alpha magically appears on the other side of equation, and the mu, the static coef, appears inside the tan.
you got confused
 
  • #3
LCSphysicist said:
μμμH=sin(α)cos(α)=μH=tan(α)

α=tan(0,7)

α=0,012(degrees)

The alpha magically appears on the other side of equation, and the mu, the static coef, appears inside the tan.
you got confused
So I have calculated the angle wrong, honestly it look odd to begin with. I'll have to try that again. But the equation is right tho? I should be able to get the angle through the static coefficient?
 
  • #4
sylent33 said:
So I have calculated the angle wrong, honestly it look odd to begin with. I'll have to try that again. But the equation is right tho? I should be able to get the angle through the static coefficient?
##μH = \frac {sin (\alpha)} {cos(\alpha)} = μH = tan(\alpha) ##

## \alpha = tan(0,7)##

## \alpha = 0,012 (degrees) ##


Now for the acceleration I thought of this;

First I would need to find the force that is pulling the object downhill; I did that like this

## F_d = m * g * sin(\alpha)##

## F = 0,0104 N ##

Than I would have to find friction force, that is "opposing" the downhill force;

## F_r = m * g * cos(\alpha)##
## F = 34,33 N ##
You got wrong in the red letters, the first is right if you put the values in the right place.
The second equation is wrong, you calculated the normal force, but yet rest the other thing we need to calc the friction force.
 
  • #5
LCSphysicist said:
You got wrong in the red letters, the first is right if you put the values in the right place.
The second equation is wrong, you calculated the normal force, but yet rest the other thing we need to calc the friction force.
Okay, I've read through your post carefully and this is what I have to say. by " the first is right if you put the values in the right place. " I would think you are implying that the question is correct, the result cannot be correct since the angle is wrong. For my equation on the friction force your right, but the mistake was made when i was putting it in latex. I actually calculated like this

$$ F_r = μH * m * g * cos(\alpha) $$

Which is,after checking also wrong I was putting in the wrong coefficient, i was susposed to put the coeffient of friction. I'll try to solve the problem with the angle first, thank you for the help sir!
 
  • #6
Okay so I have looked up in a few mathbooks and this is what I've come up with

## μH = tan(\alpha) ##
Now to get alpha we do this

##\alpha = arctan(0,7) = 35(degrees) ##

Now I calculate the downhill force

##F_d = 28,13 N## (hopefully that is right)

##F_r = 22,09## (with using the static coefficent 0,55)

Now since these forces work in opossing directions we need to subtract them

## F = F_d - F_r = 22,09 N ##

Now to get the acceleration we can use this formula ;

## F = m * a ##

## a = \frac F m ##

a = 1,208 m/s^2

Would you say that this is correct?
 
  • #7
A couple of typos in writing your post...
sylent33 said:
using the static coefficent
sylent33 said:
F=Fd−Fr=22,09N
But your final answer looks right.
 
  • #8
And what about the accelration, can I calculate the acceleration like this?
 
  • #9
sylent33 said:
And what about the accelration, can I calculate the acceleration like this?
That is what I am saying is right, your answer of 1.2 m/s2.
But before that you wrote
sylent33 said:
F=Fd−Fr=22,09N
Instead of F=Fd−Fr=28,13N-22,09N = etc.
 

Related to Determining the angle while dealing with friction on an inclined plane

1. How does friction affect the angle on an inclined plane?

Friction is a force that acts against motion, so it will decrease the angle of an object on an inclined plane. This is because the force of friction acts in the opposite direction of the object's motion, making it more difficult for the object to move up the incline.

2. How do you calculate the angle of an inclined plane with friction?

To calculate the angle of an inclined plane with friction, you can use the formula tan(θ) = μ, where μ is the coefficient of friction. This means that the angle is equal to the inverse tangent of the coefficient of friction.

3. How does the weight of an object affect the angle on an inclined plane with friction?

The weight of an object has a direct impact on the angle of an inclined plane with friction. The heavier the object, the greater the force of gravity pulling it down the incline. This means that a heavier object will require a larger angle to overcome the force of friction and move up the incline.

4. Can the angle of an inclined plane with friction be greater than 90 degrees?

No, the angle of an inclined plane with friction cannot be greater than 90 degrees. This is because at 90 degrees, the object is perpendicular to the surface and there is no component of the weight acting parallel to the surface, meaning there is no force to overcome the force of friction.

5. How can the angle of an inclined plane with friction be reduced?

The angle of an inclined plane with friction can be reduced by decreasing the coefficient of friction or by increasing the force applied to the object. This can be achieved by using a smoother surface or by adding a lubricant to reduce the frictional force. Additionally, increasing the force applied to the object, such as by using a steeper incline or a heavier object, can also reduce the angle needed to overcome the force of friction.

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