# Homework Help: I'm sorry i still don't know how to factor

1. Feb 15, 2008

### gillyr2

1. The problem statement, all variables and given/known data
Consider a lawnmower of weight "w" which can slide across a horizontal surface with a coefficient of friction "u". In this problem the lawnmower is pushed using a massless handle, which makes an angle "0" with the horizontal. Assume that F(h), the force exerted by the handle, is parallel to the handle.

Take the positive x direction to be to the right and the postive y direction to be upward.

2. Relevant equations
n-w-F(h)*sin(0) = 0 = Fx

-F(h)cos(0)+F(f)= 0 = Fy

0 = pheta
solving for F(h)
n - normal force
w - weight
F(f) friction

3. The attempt at a solution

(n-w-F(f))/(-sin(0)+cos(0))

the correct answer does not involve F(f) and n.

HELP!!!

Last edited: Feb 15, 2008
2. Feb 15, 2008

### gillyr2

can anyone help? am i suppose to use the friction coeffeicient formula?

u(k)=F(f)/n???

3. Feb 15, 2008

### kamerling

What are you supposed to calculate?

4. Feb 15, 2008

### gillyr2

F(h) the

5. Feb 15, 2008

### Cynapse

At a guess you are trying to find the total force forward contributing towards the forward motion. Either this or the values based on limiting equilibrium. Heres how you do em.

When pushing down on the handle force is F
The horizontal force going towards pushing it forward is Fcos[a] where a is the angle formed between the handle and the ground.
The vertical force down caused by F is Fsin[a]
The total vertical force down is Fsin[a]+mg
This must be the reaction force so force f due to friction is
f = alpha(Fsin[a]+mg)
so total force T pushing the mower will be
Total Force Forward = Fcos[a] - [alpha(Fsin[a]+mg)]

On the limit of eqm Force Forward = 0 so
Fcos[a] = alpha(Fsin[a]+mg)

6. Feb 15, 2008

### gillyr2

this is actual physics...with variables according to my professor

7. Feb 15, 2008

### gillyr2

didn't work sorry it was wrong. but i corrected the variables in the problem. i just noticed that they weren't excluded. my apologies

8. Feb 15, 2008

### gillyr2

The solution for F(h) has a singularity (that is, becomes infinitely large) at a certain angle 0critical. For any angle , 0>0critical the expression for F(h) will be negative. However, a negative applied force F(h) would reverse the direction of friction acting on the lawnmower, and thus this is not a physically acceptable solution. In fact, the increased normal force at these large angles makes the force of friction too large to move the lawnmower at all.

Find an expression for tan(0critical).

9. Feb 15, 2008

### gillyr2

found them both :D thats 1/u