# 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.

July 25, 2021 | education

| **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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