# Algebraic Dilemma

• Oct 4th 2009, 10:14 AM
WhoCares357
Algebraic Dilemma
Quote:

In the figure a fastidious worker pushes directly along the handle of a mop with a force F. The handle is at an angle θ = 31.33° with the vertical, and 0.42 and 0.32 are the coefficients of static and kinetic friction between the head of the mop and the floor. Ignore the mass of the handle and assume that all the mop's mass m = 0.50 kg is in its head. If the mop head moves along the floor with a constant velocity, then what is F?
I found the force to be 6.36N. I found it using $\mu_k*(mg+Fcos(\theta))=Fsin(\theta)$.

Quote:

Show that if θ is less than a certain value θ0, then F (still directed along the handle) is unable to move the mop head. Find θ0.

I setup the equation
$\mu_s*(mg+Fcos(\theta))-Fsin(\theta)=0$.

Now I can't find a way to solve for $\theta$.
• Oct 4th 2009, 11:39 AM
skeeter
solve it using technology ... even if you solve for either $\sin{\theta}$ or $\cos{\theta}$ by hand, you'll still need a calculator to find $\theta$ .
• Oct 4th 2009, 11:49 AM
Provoke
Quote:

Originally Posted by skeeter
solve it using technology ... even if you solve for either $\sin{\theta}$ or $\cos{\theta}$ by hand, you'll still need a calculator to find $\theta$ .

Don't bother, it has an infinity of solutions, he did something wrong, I'm trying to figure out what wait a sec...
.
.
.
Ok, I'll trust you that you got F right. Say its 6.36N, the driving force that pushes the mop forwards is the x component of F:

6.36N x sin(31.33) = 3.3N

so the equation is:

6.36N x sin(teta) = sum of sheer forces (static and kinetic forces)

When the 2 equal each other you are static, the mop doesn't move.

It makes sense... the more you decrease teta...the more you decrease the x component of F thats your driving force so you need to push harder on the mop. When teta is zero you are pushing into the ground so even if you suply an infinite force you will not move the mop.

cheers,

Nuke
• Oct 4th 2009, 01:44 PM
WhoCares357
Quote:

Originally Posted by Provoke
Don't bother, it has an infinity of solutions, he did something wrong, I'm trying to figure out what wait a sec...
.
.
.
Ok, I'll trust you that you got F right. Say its 6.36N, the driving force that pushes the mop forwards is the x component of F:

6.36N x sin(31.33) = 3.3N

so the equation is:

6.36N x sin(teta) = sum of sheer forces (static and kinetic forces)

When the 2 equal each other you are static, the mop doesn't move.

It makes sense... the more you decrease teta...the more you decrease the x component of F thats your driving force so you need to push harder on the mop. When teta is zero you are pushing into the ground so even if you suply an infinite force you will not move the mop.

cheers,

Nuke

I'm confused about what the right side of your equation is. Would it not just be static forces? Why would you add kinetic to it? If you just use static forces then you get the equation I came up with.
• Oct 4th 2009, 01:48 PM
skeeter
Quote:

Originally Posted by WhoCares357
I'm confused about what the right side of your equation is. Would it not just be static forces? Why would you add kinetic to it? If you just use static forces then you get the equation I came up with.

• Oct 4th 2009, 02:09 PM
WhoCares357
Quote:

Originally Posted by skeeter

I used a graphing calculator to get theta (finding intersection). I got the answer of 40.14 degrees. However the online system is telling me that I'm wrong. I retyped the equations a few times to make sure it wasn't that kind of error.
• Oct 4th 2009, 02:16 PM
skeeter
Quote:

Originally Posted by WhoCares357
I used a graphing calculator to get theta (finding intersection). I got the answer of 40.14 degrees. However the online system is telling me that I'm wrong. I retyped the equations a few times to make sure it wasn't that kind of error.

how "sensitive" is online system ? I get 40.146536 degrees
• Oct 4th 2009, 02:24 PM
WhoCares357
Quote:

Originally Posted by skeeter
how "sensitive" is online system ? I get 40.146536 degrees

It allows a 3% margin of error.

Edit: I tried using the kinetic coefficient, but that answer was wrong, too.
• Oct 5th 2009, 07:31 AM
Provoke
Quote:

Originally Posted by skeeter

Fine? ok lol. Than why is mg inside the equation? The sum of sheer forces equal the x component of F which are both along the x axis...mg is along the y axis it has nothing to do there.
• Oct 5th 2009, 08:25 AM
skeeter
Quote:

Originally Posted by Provoke
Fine? ok lol. Than why is mg inside the equation? The sum of sheer forces equal the x component of F which are both along the x axis...mg is along the y axis it has nothing to do there.

you don't know much about frictional forces, do you?

$mg + F\cos{\theta} = N$ , the normal reaction force

$
f_k = \mu_k \cdot N
$

$
f_s \le \mu_s \cdot N
$