A truck with gros weight 26000 pounds is parked with a 10 degree slope. asume the only force to overcome is that due to gravity find the force to keep the truck to keep from rolling down the hill find the force perpendicular to the hill.

F = Px = 20105.7 [N]

Fn = Py = 114055,6

Step-by-step explanation:

The truck weight 26000 pounds

1 Kg = 2.2 pounds then truck weight 26000/2.2

11818.18 kg

Then m*g = 11818,18 kg * 9.8 m/s² m*g = 115816 [N]

The truck is parked therefore there is not movement

∑ Fₓ = 0 and ∑y = 0

∑Fy = Py - Fn = 0 Py = P*cos10° ⇒ Py = 115816 * 0.9848

Py = 20105.65 [N]

and normal force is Fn = Py = 114055,6 [N]

∑Fx = m*a = 0 = Px - F

Px = P * sn 10° ⇒ Px = 115816 * 0.1736 ⇒ Px = 20105,7 [N]

And the force to keep the truck parked is:

F = Px = 20105.7 [N]