The coefficient of static friction between the m = 2.60 kg crate and
the 35.0° incline of Figure P4.41 is 0.290. What minimum force  must
be applied to the crate perpendicular to the incline to prevent the
crate from sliding down the incline?

Whew, this should be easier to imagine!
(Just been to the Science-Astronomy page. Just burned me brain
imagining them burstars, galactic plane, Myr-periods, Oort Cloud.
Zeez, t'was Chaos Cloud, now Oort Cloud. Maybe if I did not escape
from that page, I'd encounter Gungbung Clouds, whatever they are.)

So Let me imagine that Figure P4.41. (It should be in a book or
something in this planet of ours.) Maybe the 2.60 kg crate is not
being controlled by a rope being pulled by a pulley or hand. Maybe the
crate is left by itself to slide down the incline. And if the crate
tends to slide down, a force F, perpendicular to the incline, must be
applied on the crate to stop the motion. And you are asking what
minimum F is needed.

Well, without F yet, only the friction force due to the component
perpendicular to the incline of the crate's weight is there to prevent
the sliding down of the crate. And the force causing the motion is the
component parallel to the incline of the crate's weight.

Crate's weight = m*g = (2.6 kg)*(9.8 ft/sec/sec) = 25.48 newtons
Weigth's component perpendicular to incline = (25.48)cos(35deg) = 20.87 newtons
Available friction force = (0.290)(20.87) = 6.05 newtons.
Sliding force = (25.48)sin(35deg) = 14.61 newtons

Umm, the sliding force is a lot greater than the friction force
fighting it. The crate will slide! A giant forefinger is needed to
press the crate gently to prevent the crate from sliding down.

14.61 minus 6.05 = 8.56 newtons  <---to be neutralized.

friction force = (coefficient of friction)*(perpendicular force)
So,
8.56 = (0.290)*(F)
F = 8.56 / 0.290 = 29.52 newtons.

Therefore, the giant needs to exert only enough force perpendicular to
the incline of 29.52 newtons to stay the crate.