# Find the magnitude of a force

[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. $$\Sigma$$$$\vec{Fx}$$ = max

3. $$\Sigma$$$$\vec{Fy}$$ = may

4. fk = $$\mu$$kn

5. $$\vec{F}$$ = $$\sqrt{Fx^2+Fy^2}$$

## The Attempt at a Solution

$$\Sigma$$Fx = -Fcos53.1 + fk = 0

$$\Sigma$$Fy = N + Fsin53.1 - w = 0

Fcos53.1 = fk

--> Fcos53.1 = $$\mu$$kn

N = w-Fsin53.1

---> Fcos53.1 = $$\mu$$k(w-Fsin53.1)

--> Fcos53.1 = $$\mu$$kw-$$\mu$$kFsin53.1

--> Fcos53.1 + $$\mu$$kFsin53.1 = $$\mu$$kw

--> F(cos53.1 + $$\mu$$ksin53.1) = $$\mu$$kw

--> F = $$\mu$$kw / (cos53.1 + $$\mu$$ksin53.1)

--> N = Fcos53.1/$$\mu$$k

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.

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Tom Mattson
Staff Emeritus
Gold Member
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.

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)