Help with hard problem about friction

[newcool]

Hi, I need help with the following problem:

You are pushing a mop of mass m with a force P at an angle theta. The coefficiant of friction is U_k .

Find P so that the mop will start moving in terms in \theta,U_k ,m,g .

I solved this part and got:


P = \frac {(U_k *mg )}{( \sin\theta- U_k*\cos\theta)}

Now, for part 2 I have to find the minimum angle \theta for which it will be impossible for me to push the mop in terms of \theta,U_k ,m,g,P ..

Like at 90 degrees It is impossible to push the mop. Any help on part 2 would be appreciated.

Thanks

[Leong]

is theta relative to to the vertical or to the horizontal ?
if it is relative to the horizontal, i got 45 degree. Pcos(theta) = miu*N

[newcool]

Theta is relative to the Vertical. Can you explain how you got 45?

Thanks

[Leong]

45 degree is wrong, sorry. i think you have got the right anwer ninety degree.

[newcool]

The answer is below 90 degrees. From theta to 90 it will not move.

[Leong]

if theta is below ninety degree if it won't move, i don't know how the p found in the first part can move the mop.

[newcool]

P is the force required for it to move at the angle Theta. The 2nd part of the problem is to find an angle theta such that any amount of P would not be able to move the mop.

My guess is that theta is somewhere around 80 degrees, however I am looking for the true solution.

[Leong]

sorry, can't help you with that !

i found a similar question in my book. it introdueces both \mu_s for coefficient of static friction and \mu_k as coeffiecient of kinetic friction.

if the force P can't move the mop, then Psin\theta<\mu_s*N; then
Psin\theta<\mu_s*N
Psin\theta<\mu_s*(mg+Pcos\theta)
sin\theta-\mu_s*cos\theta<\frac{\mu_s*mg}{P}
\sqrt{1+\mu_s^2}*sin(\theta-tan^{-1}\mu_s)<\frac{\mu_s*mg}{P}
\theta<sin^{-1}\frac{\mu_s*mg}{P*\sqrt{1+\mu_s^2}}+tan^{-1}\mu_s
so \theta_{min}=sin^{-1}\frac{\mu_s*mg}{P*\sqrt{1+\mu_s^2}}+tan^{-1}\mu_s so that the mop will move.

[newcool]

thank you Leong, for the help
Unfortunately, it it a little too complex and there is no U_s in the problem.

The correct answer is that \tan\theta > U_k

I am still looking for how to arrive at this answer.

[Leong]

to move a thing in a frictional surface, don't we need to overcome the static friction first to get started to move, we use uk to relate the friction when it is in motion, a little weird i think for the answer to have uk.

[newcool]

I found the answer. It was actually very simple

From the answer I got for part 1:


P = \frac {(U_k *mg )}{( \sin\theta- U_k*\cos\theta)}

We can say that \sin\theta- U_k*\cos\theta > 0

Working it out you get that \frac{\sin\theta}{\cos\theta} > U_k

And then you get that \tan\theta > U_k