Determining the angle while dealing with friction on an inclined plane

[sylent33]

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.

 

[LCSphysicist]

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

 

[sylent33]

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?

 

[LCSphysicist]

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.

 

[sylent33]

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!

 

[sylent33]

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?

 

[haruspex]

A couple of typos in writing your post...

using the static coefficent

F=Fd−Fr=22,09N

But your final answer looks right.

 

[sylent33]

And what about the accelration, can I calculate the acceleration like this?

 

[haruspex]

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

F=Fd−Fr=22,09N

Instead of F=Fd−Fr=28,13N-22,09N = etc.