# On an inclined plane with a length of 13m and a height of 5m lies a load weighing 10kg. The friction coefficient is 0.2.

On an inclined plane with a length of 13m and a height of 5m lies a load weighing 10kg. The friction coefficient is 0.2. What force must be applied along the plane to pull off the load. The movement is considered uniform.

AB = 13 m – triangle hypotenuse
AC = 5 m – opposite leg
BC – adjacent leg, we find by the Pythagorean theorem = 12 m
m = 10 kg
m = 0.2
F -?
We write Newton’s second law in projection on the coordinate axis:
Ox: 0 = F – m * N – m * g * sin (a), N – support reaction force
Oy: 0 = N – m * g * cos (a)
F = m * N + m * g * sin (a)
N = m * g * cos (a)
F = m * m * g * cos (a) + m * g * sin (a) = m * g * (m * cos (a) + sin (a))
cos a = BC / AB
sin a = AC / AB
cos a = 5/13
sin a = 12/13
F = 10 * 9.8 * (0.2 * 5/13 + 12/13) = 98 N

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