Calculating Force Required to Push a Window Scrubber Up a Vertical Window

[clope023]

[SOLVED] find the magnitude of a force

Homework Statement

A window washer pushes his scrub brush up a vertical window at constant speed by applying a force \vec{F} as shown in the figure. The brush weighs 12.1 N and the coefficient of kinetic friction is 0.110.

Homework Equations

1. \Sigma\vec{F} = m\vec{a}

2. \Sigma\vec{F<span style="font-size: 9px">x}</span> = max

3. \Sigma\vec{F<span style="font-size: 9px">y}</span> = may

4. fk = \mukn

5. \vec{F} = \sqrt{F<span style="font-size: 9px">x^2+F<span style="font-size: 9px">y^2}</span></span>

The Attempt at a Solution

\SigmaFx = -Fcos53.1 + fk = 0

\SigmaFy = N + Fsin53.1 - w = 0

Fcos53.1 = fk

--> Fcos53.1 = \mukn

N = w-Fsin53.1

---> Fcos53.1 = \muk(w-Fsin53.1)

--> Fcos53.1 = \mukw-\mukFsin53.1

--> Fcos53.1 + \mukFsin53.1 = \mukw

--> F(cos53.1 + \muksin53.1) = \mukw

--> F = \mukw / (cos53.1 + \muksin53.1)

--> N = Fcos53.1/\muk

my answer was F = 1.9N, which was wrong and I was so sure I did everything right.

if anybody can offer any help it would be greatly appreciated.

 

[quantumdude]

The frictional force points in the y direction opposite the motion of the scrubber head, and the normal force points in the x direction, normal to the wall (that's why it's called the normal force!).

You've got them backwards.

 

[clope023]

The frictional force points in the y direction opposite the motion of the scrubber head, and the normal force points in the x direction, normal to the wall (that's why it's called the normal force!).

You've got them backwards.

I just realized that now, it seems the equations will turn out to be

mg + un = Fsintheta

with n = Fcostheta

eventually F = mg/(sintheta-ucostheta)

thanks for the heads up.