# Rough incline

1. Mar 21, 2006

### eok20

i am stuck on the following problem: a particle of weight w is on an inclined plane that makes an angle of t with the horizontal. the coefficient of static friction is 2*tan(t). find the minimum horizontal force h acting transverse to the slope of the plane that will cause the particle to move.

i found the normal force to be w*cos(t) so the force of friction is 2*tan(t)*w*cos(t)=2w*sin(t). i figure that this is the only force opposing h so if h=2w*sin(t) then the particle will move. however, the given answer is 3^(1/2)w*sin(t). any hints would be appreciated.

thanks.

2. Mar 21, 2006

### Chi Meson

So the applied force is horizontal? If so, one componant of the horizontal force will be along the axis perpendicular to the plane, and this will reduce the normal force.

3. Mar 21, 2006

### finchie_88

What about gravity, if this particle is on a plane, then gravity will also be present aswell as the frictional force.

Last edited by a moderator: Mar 21, 2006
4. Mar 21, 2006

### Hootenanny

Staff Emeritus
Basically, you have the following components;

$$F_{gravity}\sin (t) + F_{resistive} = h\cos (t)$$

Now as the h force is horizontal (as Chi says) it will affect the normal reaction force and hence the frictional force. So you need to resolve h so that it is perpendicular to the inclined plane and add this to the resolved gravitational force to aquire R.

5. Mar 21, 2006

### Chi Meson

Well the weight, w, is the gravitational force. I'm not sure if the OP took the parallel component of the weight into consideration or not.

6. Mar 21, 2006

### Hootenanny

Staff Emeritus
I don't think OP took into account the 'resistive' gravitational force, and as you say they didn't take into account the component of h, adding to the normal reaction force. It is an unusual question, I've never seen co-efficents for friction given as trig functions before.

7. Mar 21, 2006

### eok20

i don't see how there will be a component along the axis perpendicular to the plane because the force is perpendicular to the slope of the plane. perhaps i did a poor job explaining the problem. the problem can be found at http://www.feynmanlectures.info/exercises/weight_on_rough_incline.html

additionally i don't see where a resistive gravitational force comes into play because wouldn't that entail that if the plane were frictionless the particle would slide to a side of the plane as well as down it?

8. Mar 22, 2006

### Hootenanny

Staff Emeritus
Sorry, I miss interpreted the question.

Last edited: Mar 22, 2006
9. Mar 22, 2006

### Staff: Mentor

So far, so good. That's the maxium value of the static friction force.

No. There are three forces acting on the particle parallel to the surface of the plane: the friction, of course, which opposes any other force on the particle; the component of the weight acting down the plane; and the sideways force h. The particle will begin to slide when the sum of the latter two forces exceeds the maximum static friction. Note that those forces are perpendicular to each other.

10. Mar 22, 2006

### eok20

That make sense but I'm still not getting 3^(1/2)w*sin(t). This is what I did:

the component of the weight acting down the plane is w*sin(t) so I solved w*sin(t) + h = 2w*sin(t) for h=w*sin(t)

11. Mar 23, 2006

### Hootenanny

Staff Emeritus
The two forces (h and g) are perpendicular, so you need to resolve them into a single force. If you consider the two forces making a right triangle, then the resultant would be the hypotenues. So;

Force required for movement;

$$2w\sin\alpha$$

This is the hypotenues. So, applying pythag ($a^2 = b^2 + c^2$) gives;

$$h^2 + (w\sin\alpha)^2 = (2w\sin\alpha)^2$$

Now all you need to do is solve for h. Hope this helps.

-Hoot

12. Mar 23, 2006

### Staff: Mentor

Hoot corrected your error, but let me repeat. Forces are vectors and must be added as vectors. As I pointed out earlier, these forces are perpendicular to each other: find their resultant.