# Another Mechanics Question

1. Apr 27, 2008

### DigitalSpark

1. The problem statement, all variables and given/known data

Okay heres the first question which I done:

"Question 5: A body of mass 2kg is held in limiting equilibrium on a rough plane inclined at 20 degrees to the horizontal by a horizontal force X. The coefficient of friction between the body and the plane is 0.2. Modelling the body as a particle find X when the body is on the point of slipping:

a) up the plane.
b) down the plane."

It's the next question (which builds up on 5) which I'm having difficulty with.

"Question 6: The force X in Question 5 is replaced by a force Y at an angle of 45 degrees to the horizontal. Find Y when the body is on the point of slipping down the plane."

2. Relevant equations

3. The attempt at a solution

Okay the answer in the book says 3.67N when I get 5N. I think I'm right but I can't be sure.

Heres my method:

Rn=mgcos20 + Ysin65 ---- Equation 1

Ff+Ycos65=mgsin20 ---- Equation 2

$$\mu$$mgcos20+$$\mu$$Ysin65+Ycos65=mgsin20

Y($$\mu$$sin65+cos65)=mgsin20-$$\mu$$mgcos20

Y=$$\frac{mgsin20-\mu mgcos20}{\mu sin65+cos65}$$ $$\approx$$5N
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 27, 2008

### alphysicist

Hi DigitalSpark,

I believe when they said force Y is 45 degrees to the horizontal they meant +45 degrees, so force Y has an upwards vertical component. (Perhaps they would have specified -45 degrees the other way.)

When Y is 45 degrees above the horizontal I get Y=3.67N; when it is 45 degrees below the horizontal I get Y=5N.

3. Apr 27, 2008

### DigitalSpark

Okay but isn't Y 65 degrees to the plane?? What am I missing?

4. Apr 27, 2008

### alphysicist

When the force Y is pointing 45 degrees below the horizontal, it is 65 degrees to the plane; in which case the force Y is pushing the box into the plane (increasing the normal force).

When it is 45 degrees above the horizontal, it is 25 degrees to the plane, and pushing the box away from the plane (decreasing the normal force).

5. Apr 27, 2008

### DigitalSpark

Okay so I just assume that the force is applied anti-clockwise as in with most Trig angles??

6. Apr 27, 2008

### alphysicist

I don't think I have explained it to well with text, so I have uploaded a quick image to show you the two choices for vector Y that I mean.

http://img210.imageshack.us/my.php?image=70572227cq8.jpg

I think you were choosing the top one (that points downward and to the right) which points 45 degrees below the horizontal, but the book wants you to choose the one that points upward and to the right.

I'm not sure what you mean by the force is anti-clockwise; you break up the forces into components, and their directions indicate whether they are positive or negative.

EDIT: Oh, you must mean that the positive angles are anti-clockwise. That is a usual convention.

Last edited: Apr 27, 2008
7. Apr 27, 2008

### DigitalSpark

Yeah I choose the one that is points downward to the right. By anticlockwise, I mean the angle the line makes with the horizontal. Going anti-clockwise until it's 45 degrees.

But I simply added a 45 degree angle to X force. My problem is understanding the question I guess, where am I going wrong??

Should I ALWAYS make sure the angle is anti-clockwise and I can't go wrong??

8. Apr 27, 2008

### alphysicist

Unless they give you some indication as to one way or the other in a specific problem, then I would follow the "positive angles are counterclockwise from the +x axis" convention.

9. Apr 27, 2008

### DigitalSpark

Okay thanks.

10. Apr 27, 2008

### DigitalSpark

One problem though. In the the a) part of question 6 it asked:

"Question 6: The force X in Question 5 is replaced by a force Y at an angle of 45 degrees to the horizontal. Find Y when the body is on the point of slipping

a) up the plane."

And I basically used (-45 degrees). And I got the answer in the book. How do you explain that??