What angle does the ramp make with the ground Static friction

[ryan]

A box of mass 32.0 kg sits on a horizontal steel ramp 3.4 m long with a coefficient of static friction of 0.30 between the ramp and the box. The end of the ramp is slowly lifted until the box begins to slide down the ramp. What angle does the ramp make with the ground when this happens?

What i have so far:
In the x-direction there is a frictional force of -.30 and a force of mgsin(&theta). How can i determine if the box will move with the angle. F - fr = ma?

 

[HallsofIvy]

The box will start to move as soon as the net force is greater than 0: as soon as mg sin(θ)= .30.

By the way, you need a ";" after the &theta.

 

[sridhar_n]

Hi

I don't agree with Hallsof Ivy. The situation is similar to a box lying on an iclined plane and finding the angle of inclination at which the box will start sliding. As the ramp is lifted, the box stays in position as long as &mu*mg*cos&theta = mg*sin&theta.(&theta being the angle of inclination).Which means that the box will slide down the slope as soon as the angle that the ramp makes with the horizontal exceeds:

&theta, where tan&theta = &mu, where &mu is the coefficient of static friction between the ramp and the box = 0.3.

This angle is called as the "[BAngle of Repose[/B]"

Sridhar

 

[Antepolleo]

I'm not quite sure how you got that friction force, but I'm assuming that you chose the horizontal plane as your x-direction. This is an example of a problem that is greatly simplified if we choose our coordinate axes to be in a different orientation than usual. It is always helpful to have the greater number of forces pointing along the axes, so let's call the positive x direction to be sloped down the ramp, and the positive y direction to be oriented with the normal force.

This means that the only force that is not acting along a coordinate axes would be the force of gravity, making our lives much more simple.

Now back to the meat of the problem. The equation that defines the force of static friction would be

f_s [<=] &mu;_sn

with &mu;_s being the coefficient of static friction and n being the normal force. Because we are looking for the angle in which the block just starts to slip, we can say

f_s = &mu;_sn

Now we draw the free body diagram. Unfortunately I am unaware of a program in which I can do this easily, so I'll describe it the best I can.

In the y direction, we have the normal force pointing straight upwards. Due to our tilted coordinate axes, we have a vertical and a horizontal component of gravity, so we have a vertical component of gravity pointing downwards. Using some geometric laws we can find that the angle that gravity makes with the horizontal is actually just &theta;, our original slope of the ramp. Another way to picture is to imagine that &theta; is zero... in that case, the ramp would be level with the ground, cos(&theta;) = 1, and gravity would point straight down. Therefore:

[sum] F_y = n - mgcos(&theta;)

In the x direction, we have the horizontal component of gravity pointing down the ramp, and the force of static friction pointing up the ramp, hence:

[sum]F_x = mgsin(&theta;) - f_s

Since the block will not be changing velocity (we are looking for the angle in which it is juuuust holding on for dear life) we can set the summation of the forces equal to zero in both directions.

0 = mgsin(&theta;) - f_s

0 = n - mgcos(&theta;)

Recall that

f_s = &mu;_sn

so

0 = mgsin(&theta;) - &mu;_sn

Using the forces in the y direction, we can solve for the normal force:

n = mgcos(&theta;)

We're almost here. Let's plug that back into our equation for the forces in the x direction:

0 = mgsin(&theta;) - &mu;_smgcos(&theta;)

Right away you should see that m and g cancel out of the equation, so our answer is indepentant of the mass of our object and of the acceleration of gravity:

0 = sin(&theta;) - &mu;_scos(&theta;)

Now we bring the sine function to the other side, and divide both sides by the cosine function:

sin(&theta;) / cos(&theta;) = &mu;_s

or

tan(&theta;) = &mu;_s

So we now have our answer for theta, which is the critical angle in which the block gives way:

&theta; = arctan(&mu;_s)

Hope this helps. Any questions are welcome.

 

[sridhar_n]

My previous reply, more organized...

Hi, since my previous reply did not contain thetas...i am re formatting my reply here...

The situation is similar to a box lying on an iclined plane and finding the angle of inclination at which the box will start sliding. As the ramp is lifted, the box stays in position as long as &mu*mg*cos&theta; = mg*sin&theta;.(&theta; being the angle of inclination).Which means that the box will slide down the slope as soon as the angle that the ramp makes with the horizontal exceeds:

&theta;, where tan&theta; = &mu;, where &mu; is the coefficient of static friction between the ramp and the box = 0.3.

This angle is called as the "Angle of Repose"

Sridhar

 

[Antepolleo]

Heh, maybe I should write a book.

 

[sridhar_n]

What makes you say that Antepolleo ?

Heh, maybe I should write a book.

What makes u say that?

 

[Antepolleo]

Originally posted by sridhar_n
What makes u say that?

I think I may have went into too much detail, is all. No reason to get excited.

 

[Antepolleo]

I wrote all that and he didn't even thank me. I hope his naughty bits experience a high degree of kinetic friction.

 

[FrostScYthe]

I wrote all that and he didn't even thank me. I hope his naughty bits experience a high degree of kinetic friction.

I'll thank you, that really helped