Question

A mop is pushed across the floor with a force FF of 50.0 N,
at an angle of θ = 50.0°. The mass of the mop head is 3.75 kg.

Calculate the acceleration of the mop head if the coefficient
of kinetic friction between the head and the floor is μk= 0.400.

Answer 1

The image of the situation:
http://i.imgur.com/ctpvhx6.jpg

Answer 2

Am I doing it correct?

The sum of forces in the x:
\(\color{#000000}{ \large \displaystyle F_{k ,~x}+F_{{\rm push} ,~x} =ma_x }\)

Friction is proportional to normal force:
\(\color{#000000}{ \large \displaystyle F_{k}=\mu_kN= \mu_kmg }\)

So, the friction-force in the x would be:
\(\color{#000000}{ \large \displaystyle F_{k,~x}= \mu_kmg\cos\theta }\)

And the force of push in the x is:
\(\color{#000000}{ \large \displaystyle F_{{\rm push} ,~x}= F_{{\rm push} } \cos \theta }\)

Therefore: (the force of friction is negative, since it counteracts the relative motion. That is why I got the minus there.)
\(\color{#000000}{ \large \displaystyle -\mu_kmg\cos\theta + F_{{\rm push} } \cos \theta =ma_x }\)

With these particular numbers I have:
\(\color{#000000}{ \large \displaystyle -(0.4)(3.75)(9.80)\cos(50) + 50 \cos(50) =3.75~a_x }\)

So, the acceleration in the x (and the only acceleration, since the object is obviously not accelerating vertically) is:
\(\color{#000000}{ \large \displaystyle a_x=a=\frac{50 \cos(50)-(0.4)(3.75)(9.80)\cos(50)}{3.75} \approx 6.05077 }\)

Answer 3

so i think how i see your calculi that you are right - good work

bye

Answer 4

It seemed reasonable to me too, but this answer turned out to be incorrect-:(

Answer 5

why what will be the right answer ?

Answer 6

I don't know. That's what I am trying to find out.
I took all of the forces that act in the x-direction into account, but for some reason this answer is incorrect.

Answer 7

@Michele_Laino

Answer 8

the friction force \(R\) is:
\[R = \mu \left( {mg + F\sin \theta } \right)\]
so the requested acceleration, is:
\[a = \frac{{F\cos \theta - R}}{m}\]

Answer 9

\(\color{#000000}{ \large \displaystyle R= \mu_k(mg + F \sin \theta) }\)

\(\color{#000000}{ \large \displaystyle a=\frac{F \cos \theta-\mu_k(mg + F \sin \theta) }{m} }\)

\(\color{#000000}{ \large \displaystyle a=\frac{50 \cos 50-0.4(3.75\cdot 9.80 + 50 \sin \theta) }{3.75} }\)

Like this?

Answer 10

Oh, forgot to plug in sin(50)...

Answer 11

yes!

Answer 12

thanks, I will try that.
One question, tho, how did you come up with this formula for R?

Answer 13

since the magnitude of R is proportional to the orthogonal pressure exerted by the mop

Answer 14

orthogonal with respect to the horizontal Surface, of course

Answer 15

Well, or, in two dimensions just perpendicular.
So, that would mean sine, not cosine.
And proportion is the friction coefficient.

Answer 16

yes!

Answer 17

That starts to kind of making sense to me. Thanks!

Answer 18

:)