Question

A window washer pushes his scrub brush →−
up a vertical window at constant speed by applying a force F as
shown in Fig. P5.74. The brush weighs 15.0 N and the coefficient
of kinetic friction is μk = 0.150. Calculate (a) the magnitude of →−
the force F and (b) the normal force exerted by the window on the brush.

Answer 1

I don't understand your symbology →−

Answer 2

where's the picture?

any how, F=uN
brush weighs 15N , 15N=the normal force if the washer doesn't push down. If he/she did push down on the brush with a force F, it would be the cosine of the angle that is pushing down . so yeah draw a free body, and you'll see which forces are required

Answer 3

v=const => a=0

N =F(x)•cosα
F(y)•sinα=mg+F(fr)
F(y)=(mg+μN)/sinα
F=sqrt(F(x)²+F(y)²)

Answer 4

F is 208.42N normal is 125.14 N

Answer 5

F*sin(theta) = F*cos(theta) * (mew) + weight
F*sin(theta) = F*cos(theta) * (0.150) + 15

Answer 6

Fsin(theta) - 15 - (Uk)N=0
Fsin(theta) - (.150)Fcos(theta)= 15
Plug in theta and solve for F. Once you get F use equation
Fcos(theta)-N=0 to solve for N